Irving  Strin^han 


'*th. 


-4^    .C 


f 


NOTES 


ON 


ELEMENTS 


OF 


(ANALYTICAL) 

SOLID  GEOMETRY 


BY 

CHAS.  S.  VENABLE,  LL.D., 

PROFESSOR  OF  MATHEMATICS,  UNIVE-RSITY  OF  VIRGINIA. 


NEW   YORK: 
UNIVERSITY    PUBLISHING   COMPANY. 

1891. 


Math.  0o«t. 


IN  preparing  these  Notes  I  have  used  the  treatises  of  Gregory, 
Hymers,  Salmon,  Frost  and  Wostenholme,  Bourdon,  Sonnet  et 
Frontera,  Joachimsthal-Hesse,  and  Fort  und  Schlomilch. 

C.  S.  V. 


V 


COPYRIGHT,  BY 

UNIVERSITY  PUBLISHING  COMPANY 
1879. 


NOTES  ON  SOLID  GEOMETRY. 


CHAPTER  I. 

1.  WE  have  seen  how  the  position  of  a  point  in  a  plane  with  ref 
erence  to  a  given  origin  O  is  determined  by  means  of  its  distances 
from  two  axes  (Xr,  Oy  meeting  in  O.      In  space,  as  there  are  three 
dimensions,  we  must  add  a  third  axis  Oz.     So  that  each  pair  of  axes 
determines  a  plane,  CXr   and  Oy  determining  the  plane  xOy  ;    O.v 
and  O2  the  plane  xOz  ;  Oy  and  Oz  the  plane  yOz.     And  the  posi 
tion  of  the  point  P  with  reference  to  the  origin  O  is  determined  by 
its  distances  PM,  PN,  PR  from  the  zOy,  zOx,  xOy  respectively,  these 
distances  being  measured  on  lines  parallel  to  the  axes  CXv,  Oy  and 
Oz  respectively.     This  system  of  coordinates  in  space  is  called  The 
System  of  Triplanar  Coordinates,  and  the  transition  to  it  from  the 
System  of  Rectilinear  Plane  Coordinates  is  very  easy.     We  can  best 
conceive  of  these  three  coordinates   of  P  by  conceiving  O  as  the 
corner  of  a  parallelopipedon  of  which  OA,  OB,  OC  are  the  edges, 
and  the  point  P  is  the  opposite  corner,  so  that  OP  is  one  diagonal  of 
the  parallelopipedon. 

2.  If  PM  =  OA  =  a,  PN  =  OB  =  b,  PR  =  OC  =  c,  the  equations 
of  the  point  P  are  x  =  a,  y  =  b,  z  =  c,  and  the  point  given  by  these 
equations  may  be  found  by  the  following  construction  :  Measure  on 
OX  the  distance  OA  =  a,   and  through  A  draw  the    plane  PNAR 
parallel  to  the  plane   yOz.      Measure  on  Oy  the  distance  OB  =  3, 
and  draw  the  plane  PMBR  parallel  to  xOz,  and   finally  lay  off  OC 
—  c  and  dnuv  the  plane  PMCN  parallel  to  xOy.     The  intersection 
of  these  three  planes  is  the  point  P  required.      (Fig.  i.)* 

3.  The  three  axes  Or,  Oy,  Oz  are  called  the  axes  of  x}  y,  and  z 
respectively  ;    the    three  planes  xQy,   xOz,   and   yOz  are  called  the 

*. 

*  For  Figures  see  Plates  I.  and  II.  at  end  of  book. 

814023 


4  NOTES  ON   SOLID    GEOMETRY. 

planes  xy,  xz  and  yz  respectively.     The  point  whose  equations  are 
x  —  a,  x  =  b,  x  =  c  is  called  the  point  (a,  b,  c). 

4.  The  coordinate  planes  produced  indefinitely  form  eight  solid 
angles  about  the  point  O.      As  in  plane  coordinates  the  axes  Ox 
and  O>'  .cUvicieUtte  plane  considered  into  four  compartments,  so  in 
space  coordinates  the  planes  xy.   xz  and  yz  divide  the  space  con- 
•slcjttfed  'ia.;o  '.eight  Compartments — four  above  the   plane  xy,   viz.  : 
'Q-xyz,  Q-xy'z,   Q-x'y'z,   Q-x'yz  ;    and   four  below  it,   viz.  :    Q-xyz', 
Q-xy'z,   O-xfyfz',   Q-x'yz'.      By  an  easy  extension    of  the   rule  of 
signs  laid  down  in   Plane  Coordinate  Geometry,  we  regard  all  x's 
on  the  right  of  the  plane  yz  as  +  and  on  the  left  of  yz  as  —  ;  ally's 
in  front  of  the  plane  xz  as  -f-  and  those  behind  it  as  —  ;  all  z's  above 
the  plane  xy  as  +  and  those  below   it  as  —  .     We  can  then  write  the 
points  whose  distances  from  the  coordinate  planes  are  a,  b  and  c  in 
the  eight  different  angles  thus  : 

In  the  first  Octant,  Q-xyz  Pl  is              (a,  b,  c) 

In  the  second  Octant,  P2  is  (a,  —  b,  c) 

In  the  third  Octant,  P3  is  (—  a,  —  b,  c] 

In  the  fourth  Octant,  P4  is  (—  a,  b,  c) 

In  the  fifth  Octant,  P5  is  (a,  b,  —  c) 

In  the  sixth  Octant,  P6  is  (a,  —  b,  —  c] 

In  the  seventh  Octant,  I\  is  (—  a,  —  b,  —  c) 

In  the  eighth  Octant,  P8  is  (—  a,  b,  —  c). 

The  signs  thus  tell  us  in  which  compartment  the  point  falls, 
and  the  lengths  of  a,  b  and  c  give  us  its  position  in  these  compart 
ments. 

1.  Construct  the  points    I,  —2,  3  ;  o,  —  i,  2  ;  0,0,   i  ;    —  4,  o,  3. 

2.  Construct  the  points  i, — 3, —4  ;   2,— 3,  o  ;  3,  o,  —  i  ;     2,  o,  o. 

5.  The  points  M,  N  and  R  are  called  the  projections  of  P  on  the 
three  coordinate  planes,  and  when  the  axes  are  rectangular  they  are 
its  orthogonal  projections.     We  will  treat  mainly  of  orthogonal  pro 
jections.      For  shortness'  sake  when  we  speak  simply  of  projections, 
we  are  to  be  understood  to  mean  orthogonal  projections,  unless  we 
state  the  contrary. 

We  will  give  now  some  other  properties  of  orthogonal  projections 
which  will  be  of  use  to  us. 


NOTES   ON  SOLID   GEOMETRY.  5 

6    DEFINITIONS. 

The  projection  of  a  line  on  a  plane  is  the  line  containing  the 
projections  of  its  points  on  the  plane. 

When  one  line  or  several  lines  connected  together  enclose  a  plane 
area,  the  area  enclosed  by  the  projection  of  the  lines  is  called  the 
projection  of  the  first  area. 

The  idea  of  projection  may  be  in  the  case  of  the  straight  line  thus 
extended:  if  from  the  extremities  of  any  limited  straight  line  we  draw 
perpendiculars  to  a  second  line,  the  portion  of  the  latter  intercepted 
between  the  feet  of  the  perpendiculars  is  called  the  projection  of  the 
limited  line  on  the  second  line. 

From  this  we  see  that  OA,  OB  and  OC  (coordinates  rectangular) 
are  the  projections  of  OP  on  the  three  axes,  or  the  rectangular  coordi 
nates  of  a  point  are  the  projections  of  its  distance  from  the  origin  on  the 
coordinate  axes. 

7.   FUNDAMENTAL  THEOREMS. 

I.  The  length  of  the  projection  of  a  finite  right  line  on  any  plane  is 
equal  to  the  line  multiplied  by  the  cosine  of  the  angle  which  it  makes  with 
tht,  plane. 

Let  PQ  be  the  given  finite  straight  line,  xOy  the  plane  of  pro 
jection  ;  draw  PM,  QN  perpendicular  to  it ;  then  MN  is  the  projec 
tion  of  PQ  on  the  plane.  Now  the  angle  made  by  PQ  with  the  plane 
is  the  angle  made  by  PQ  with  MN.  Through  Q  draw  QR  parallel 
to  MN  meeting  PM  in  R,  then  QR  =  MN,  and  the  angle  PQR 
=  the  angle  made  by  PQ  with  MN.  Now  MN  =  QR  =  PQ  cos 
PQR.  (Fig.  2.) 

II.  The  projection  on  any  plane  of  any  bounded  plane  area  is  equal  to 
thut  area  multiplied  by  the  cosine  of  the  angle  between  the  planes. 

i°.  We  shall  begin  with  a  triangle  of  which  one  side  BC  is  parallel 
to  the  plane  of  projection.  The  area  of  ABC  =  -  BC  x  AD,  and  the 

area  of  the  projection  A'B'C'  =  -  B'C'  x  A'D.     But  B'C'  =  BC  and 

A'D'  =  AD  ccs  ADM.     Moreover  ADM  =  the  angle  between  the 
planes.    Hence  A'B'C= ABC  x  cos  angle  between  the  planes.  (Fig.  3.) 

2°.   Next  take  a  triangle  ABC  of  which  no  one  of  the  sides  is  pa 
rallel  to  the  plane  of  projection.      (Fig.  4.) 
i* 


6  NOTES  ON  SOLID   GEOMETRY. 

Through  the  corner  C  of  the  triangle  draw  CD  parallel  to  the 
plane  of  projection  meeting  AB  in  D.  Now  if  we  call  6  the  angle 
between  the  planes,  then  from  i°  A'C'D'  =  ACD  cos  6  and  B'C'D' 
=  BCD  cos  0.  .'.  A'B'D'-B'C'D'=(ABD  -  BCD)  cos  0  or  A'B'C' 
=  ABC  cos  d. 

3°.  Since  every  polygon  may  be  divided  up  into  a  number  of 
triangles  of  each  of  which  the  proposition  is  true — it  is  true  also  of 
the  polygon,  i.  e.,  of  the  sum  of  the  triangles. 

Also  by  the  theory  of  limits,  curvilinear  areas  being  the  limits  of 
polygonal  areas,  the  proposition  is  also  true  of  these. 

8.  The  projection  of  a  finite  right  line  upon    another   right  line   is 
equal  to  the  first  line  multiplied  by  the  cosine  of  the  angle  between  the 
lines. 

Let  PQ  be  the  given  line  and  MN  its  projection  on  the  line  CXv,  by 
means  of  the  perpendiculars  PM  and  QN.  Through  Q  draw  QR 
parallel  to  MN  and  equal  to  it.  Then  PQR  is  the  angle  made  by 
PQ  with  Ox,  and  MN  =  QR  =  PQ  cos  PQR.  (Fig.  5.) 

9.  If  there  be  three  points  P,  P',  P"  joined  by  the  right  lines  PP', 
PP"  and  P'P",  the  projections  of  PP"  on  any  line  will  be  equal  to 
the  sum  of  the  projections  of  PP'  and  P'P"  on  that  line.      Let  D,  D', 
D"  be  the  projections   of  the   points  P,    P',    P"   on   the  line   AB. 
Then  D'  will  either  lie  between  D  and  D"  or  D"  between  D  and 
D'.     In  the  one  case  DD"  =  DD'  +  D'D"  and  in  the  other  DD"= 
DD'  -  D"D'  =  in  both  cases  the  algebraic  sum  of  DD'  and  D'D". 
The  projection  is  -f  or  —  according  as  the  cosine  of  the  angle  above 
is  -f  or  — . 

In  general  if  there  be  any  number  of  points  P,  P',  P",  etc.,  the  pro 
jection  of  PP"'  on  any  line  is  equal  to  the  sum  of  the  projections  of 
PP',  P'P",  etc.,  or,  the  projection  of  any  one  side  of  a  closed  po 
lygonal  line  on  a  straight  line  is  equal  to  the  sum  of  the  projections 
of  the  other  sides  on  that  line. 

10.  USEFUL  PARTICULAR  CASE. 

The  projection  of  th?  radius  vector  OP  of  a  point  P  on  any  line  is 
equal  the  sum  of  the  projections  on  that  line  of  tlie  coordinates  OM,  MN, 
NP  of  the  point  P.  For  OMNP  is  a  closed  broken  line,  and 
the  projection  of  the  side  OP  on  a  straight  line  must  be  equal  to 
the  sum  of  the  projections  of  the  sides  OM,  MN,  and  NP  on  that 
line. 


AZOTES   OA>    SOLID   GEOMETRY.  7 

11.  DISTANCE  BETWEEN  Two  POINTS. 

Let  P  and  Q,  whose  rectangular  coordinates  are  (x,y,  z]  and  (x ', 
y,  z'),  be  the  two  points.  (Fig.  6.) 

We  have  from  the  right  parallelopipedon  PMNRQ  of  which  PQ 
is  the  diagonal,  PQ2  =  PM2+  MN2  +  QN2.  But  PM  =  x' -  x,  MN 

—y—y'y  NQ=  z'—z. 

Hence  PQ2  =  (x  -  xj+  (y  -y'Y  +  (*  -  z')\ 
If  one  of  the  points  P  be  at  the  origin  then  x  —  o,  y  =  o,  2=0, 
and  PQ2  =  *'2  +  /2+z"2. 

12.  TO  FIND  THE  RELATIONS  BETWEEN  THE  COSINES    OF    THE  ANGLES 

WHICH  A  STRAIGHT  LINE  MAKES  WITH  THREE  RECTANGULAR 
AXES. 

Take  the  line  OP  through  the  origin.  Let  OP  —  r,  the  angle 
POjt:  =  of,  POy  =  /3,  POz  —  y,  and  x{ ',  y' ,  z'  the  coordinates  of  P. 

Then  by  Art.   1 1,  r*  =  ,v'2+y2  +  z'-. 

But,  Art.  8,     x'  =  r  cos  a  ;  y  =  r  cos  ft  ;  z'  =  r  cos  y. 

Hence  r1  =  r2  (cos2  a  -f  cos2  ft  -\-  cos2  ^)  or 

cos2  <?  +  cos2  ft  +  cos2  y  =  i.     (i)     A  very  im 
portant  relation. 

Cos  #,  cos  /?,  cos  y  determine  the  direction  of  the  line  in  rectan 
gular  coordinates,  and  are  hence  called  the  direction  cosines  of  the  line. 
We  usually  call  these  cosines  /,  m  and  «  respectively.  So  the  equa 
tion  (i)  is  usually  written  P  +  m^  +  n?  =  i,  (i),  and  when  we  wish  to 
speak  of  a  line  with  reference  to  its  direction,  we  may  call  it  the  line 
(/,  m,  ii).  Only  two  of  the  angles  a,  ft,  y  can  be  assumed  at  pleas 
ure,  for  the  third,  y,  will  be  given  by  the  equation 


cos  y  =  ±  A/1  —  cos2  a  —  cos2  ft. 

13.  We  can  use  these  direction  cosines  also  for  determining  the 
position  of  any  plane  area  with  reference  to  three  rectangular  coordi 
nate  planes.  For  since  any  two  planes  make  with  each  other  the 
same  angle  which  is  made  by  two  lines  perpendicular  to  them  respec 
tively,  the  angles  made  by  a  plane  with  the  rectangular  coordinate 
planes  are  the  angles  made  by  a  perpendicular  to  the  plane  with  the 
coordinate  axes  respectively.  Thus  if  OP  be  the  perpendicular  to  a 
plane,  the  angle  made  by  the  plane  with  the  plane  xy  is  the  angle  y  ; 
with  xz  is  the  angle  ft ;  and  with  yz  is  the  angle  a.  So  cos  a,  cos 
ft,  cos  y,  are  called  also  the  direction  cosines  of  a  plane.  That  is,  the 


g  iVOTES   ON   SOLID    GEOMETRY. 

direction  cosines  of  a  plane  with  reference  to  rectangular  coordinates  are 
the  direction  cosines  of  a  line  perpendicular  to  this  plane. 

14.  The  relation  cos2  a  +  cos2  ft+  cos2  y  =  i  enables  us  to  prove 
an  important  property  of  the  orthogonal  projections  of  plane  areas. 
For  let  A  be  any  plane  area,  and  Ax,  Ay>  Az  its  projections  on  the 
coordinate  planes  yz,  zx  and  ;ry  respectively.     Then  Art.  7,  II.,  Az 
=  A  cos  a  ;  Ay  =  A  cos  ft  ;  Aa  =  A  cos  y. 

Squaring  and  adding  we  have 

AZ2+  A,,2+  A.9  =  A2  (cos2  a  +  cos2  ft  -f  cos2  y) 

or  A/+  A/  +  A39  =  A2. 

That  is,  the  square  of  any  plane  area  is  equal  to  the  sum  of  the  squares 
of  its  projections  on  three  planes  at  right  angles  to  each  other. 

15.  To    FIND    THE  COSINE    OF  THE    ANGLES    BETWEEN  Two  LlNES     IN 

TERMS  OF  THEIR  DIRECTION  COSINES  (cos  <*,  cos  ft,  cos  y) 
AND  (cos  a',  cos  ft',  cos  y'). 

Draw  OP,  OQ  through  the  origin  parallel  respectively  to  the  given 
lines.     They  will  have  the  same  direction  cosines  as  the  given  lines, 
and  the  angle  POQ  will  be  the  angle  between  the  given  lines.  (Fig.  7.) 
Let  POQ  =  6,  OP  =  r,  OQ  —  r',  coordinates  of  P(x,y,  z\  coor 
dinates  of  Q  (xy'z'}. 
Now  by  Art.  n, 

PQ*  =  (x-x>y  +  (y  -yj  +  (Z  -  Zj  =  X*  +  f  +  Z*  +  X*  +  /» 

2}>/  +   2ZZ'}. 


And  from  triangle  POQ, 

PQ2  =  r2  -f  r'*  —  2rr'  cos  0, 

hence     r9+  r'2-  2rr'  cos  B  =  x*  +/+  z*  +  x*  +/2+  z*  — 

-f  2yy'  +  2zz'). 

But  r*=  .r2+j'2+  ^2  and  r'2=  ,r'2  +  ya+  ^'2. 

Therefore  rr;  cos  6  =  xx'  +  yy'  +  zz  , 

a       x    x'     y     y'      z    z' 

or  cos  B  =  —.—+—.--—+-.—. 

r    r       r     r       r    r 

Hence     cos  6  =  cos  «  cos  <*'  +  cos  ^cos  ft'  +  cos  y  cos  y'     (i) 
which  we  write  cos  B  —  //'  +  ww'  +  #«'.  (2) 


NOTES   ON  SOLID    GEOMETRY.  g 

Cor.  i.  If  the  lines  are  perpendicular  to  each  other  cos  6  =  o  or 
//'+  mm'  +  nri  =  o  (3).  (3)  is  called  the  condition  of  perpendicu 
larity  of  the  two  lines  (/,  m,  «),  (/',  m',  »'). 

Cor.  2.  From  expression  for  cos  6  we  find  a  convenient  one 
for  sin2  6. 

Thus      sin2  0=1-  (//'+  mm'+  nn'Y  =  (/2+  m*+  n*}  (l'z+m* 

+  n'2)  —  (//'  +  mm'  +  nn'Y 
whence        sin2  6  =  (Irti  —  I'm}*-}-  (lnr  —  /'«)2  +  (mn'  —  m'n)\     (4) 

1  6.  To  express  the  distance  between  two  points  in  terms  of  their 
oblique  coordinates. 

Let  P  (xyz)  and  Q  (x'y'z')  be  the  two  points.      (Fig.  8.) 
The   parallelopipedon    MPQN  is    oblique.     Let    the    angle   xOy 
=  A,   xOz  =  //,  yOz  =  v,  and    the   angles  made   by  PQ  with  the 
axes  respectively  a,  ft  and  y.      Project  the  broken  line  PMNQ  on 
PQ.     This  projection  is  equal  to  TQ  itself.      Hence  we  will  have 

PQ  =  PM  cos  a  +  MN  cos  ft  +  NQ  cos  y.     (a] 

Now  project  the  broken  line  PMNQ  on  the  axes  xyz  respectively. 
We  obtain  thus  the  three  equations 

PQ  cos  a  =  PM  +  MN  cos  A.  +  NQ  cos  /*  j 
PQ  cos  ft  =  PM  cos  A  +  MN  +  NQ  cos  v  V    (b) 
PQ  cos  y  —  PM  cos  /*  +  MN  cos  v  +  NQ  ) 

Now  multiply  the  first  of  equations  (3)  by  PM,  the  second  by  MN 
and  the  third  by  NQ  and  add  them  taking  (a)  into  account  and  we 
have 

PQ2  =  PM2  +  MN2+  NQ2+  2PM  .  MN  cos  A  +  2PM  .  NQ  cos 

/*  +  2MN,NQcos  v     (c) 

or   PQ2  =  (X  -  XJ  +  (y  -/)'+  (z  -  z'Y+  2(x  -  x')  (y  -y') 
cos  /I  +  2(x—  x')(z  —  z')  cosfii  +  2(y—  -y')(z—  z')  cos  v.    (5) 


Cor.   If  one  of  the  points  as  Q  be  at  the  origin  then 

PO2  —  x*  +  y*  +  z*  -f  2xy  cos  A+  2x2  cos  jn  +  2zy  cos  v.     (6) 

17.   Direction  Ratios.     In  oblique  coordinates  the  position  of  a  line 

PM       MN        NQ 
PQ  is  determined  by  the   ratios  -j-y  ;    -^—  -  ;    ~p^-,  and  these  we 

call    direction  ratios.      We  may  name    these    /,    m,   n   respectively, 


IO  NOTES  ON  SOLID   GEOMETRY. 

taking  care  to  note  that  \ve  are  using  oblique  coordinates  and  call 
the  line  PQ,  the  line  (/,  ;//,  n}.  To  find  a  relation  among  these 
direction  ratios,  we  divide  equation  (c)  Art.  16,  by  PQ2.  We  thus 
have 

i  =  I'2  +  m*  +  ;/2  +  2//T2  cos  A  +  2ln  cos  /*  +  2mn  cos  v,  (7)  the 
desired  relation. 

1  8.  The  coordinates  of  the  point  (xyz)  dividing  in  the  ration  :  n 
the  distances  between  the  two  points  (x'y'z1)  (x"y"z")  are 

mx"  +  nx!  my"  +  ny'  mz"  +  nz' 

x  •=.  —  —  ,  y  =  —£—       —  .  z=  -  —  .      (8) 

m  +  n        J          7?i  +  n  m  +  n 

The  proof  of  this  is  precisely  the  same  as  that  for  the  correspond 
ing  theorem  in  Plane  Coordinate  Geometry. 

19.  POLAR  COORDINATES. 

The  position  of  a  point  in  space  is  also  sometimes  expressed  by 
the  following  polar  coordinates  : 

The  radius  vector  OP  =  r,  the  angle  PO0  =  6  which  the  radius 
vector  makes  with  a  fixed  axis  Oz,  and  the  angle  CO^tr—  cp  which  the 
projection  OC  of  the  radius  vector  on  a  p'ane  yQx  perpendicular  to 
O0  makes  with  the  fixed  line  Ox  in  that  plane.  (Fig.  9.) 

We  have  OC  =  r  sin  6.  Hence  the  formulae  for  transforming  from 
rectangular  to  these  polar  coordinates  are 

x  —  r  sin  8  cos  <p  } 

Y  —  r  sin  8  sin    <p\    (9) 

z  =  r  cos  8  j 


and  these  give  r1  —  x^  +.V2  + 


tan  cp  =  — 


a       *  z 

cos  8  =  —  —  — = 


(10). 


Conceive  a  sphere  described  from  the  centre  O,  with  a  radius  =  a 
and  let  this  represent  the  earth.  Then,  if  the  plane  zOx  be  the 
plane  of  the  first  meridian  and  the  axis  of  z  the  axis  of  the  earth, 

Q  — latitude,    cp  =    longitude  of  a  point  on  the  earth's   sur 
face. 


NOTES   ON   SOLID   GEOMETRY.  Ir 

20.    Distance  between  two  points  in  space  in  polar  coordinates. 

Let  P  be  (rr,  6',  q>'}  and  Q  (r,  6,  <p).  Project  PQ  on  the  plane  xy, 
MN  is  this  projection,  draw  OM  and  ON  the  projections  of  OP  and 
OQ  respectively  on  that  plane.  Through  P  draw  PR  parallel  to  MN, 
then  PR  =  MN.  (Fig.  10.) 

And  we  have 

PQ2  =  PR2  +  RQ2  =  MN2  +  (QN  -  RN)2. 

But  in  triangle  MON 

MN2  ==  OM8+  ON2-  2OM  .  ON  cos  MON, 
or     MN2  =  r'2  sin2  &  +  r8  sin2  6  —  2rr'  sin  0  sin  d'  cos  (tp  —  <p'). 

Moreover         QN  —  r  cos  0  and  RN  =  PM  =  r  cos  6'. 

Hence     PQ2=  r'2  sin2  0'  +  r*  sin2  0  -  2rr'  sin  0  sin  0'  cos  (q>  —  <p') 

+  (r  cos  0  -  r'  cos  0')* 
or 

PQ2=r'2+r2-2rr'(cos  6>cos^'  +  sin  0  sin  6'  cos  (<p  -  q>')).     (n) 


CHAPTER    II. 

INTERPRTTATION    OF    EQUATIONS. 

TRIPLANAR    COORDINATES. 

21.  LET  us  take  F  (xyy,  z]  =  o,  that  is  any  single  equation  con 
taining  three  variables  x,  y  and  z.     This  may  be  considered  as  a 
relation  which  enables  us  to  determine  any  one  of  the  variables  when 
the  other  two  are  given.     Let  these  be  x  and  j'.     So  the  equation 
may  be  written 

*=/(•*,  jOi 

in  which  we  may  attribute  arbitrary  and  independent  values  to  x  and 
y.  And  to  every  pair  of  such  values  there  is  a  determinate  point  in  the 
plane  xy ;  and  if  through  each  of  these  points  we  draw  a  line  parallel 
to  the  axis  of  z,  and  take  on  it  lengths  equal  to  the  values  of  z  given 
by  the  equation,  it  is  clear  that  in  this  way  we  will  get  a  series  of 
points  the  locus  of  which  is  a  surface,  .and  not  a  solid  since  we  take 
determinate  lengths  on  each  of  the  lines  drawn  parallel  to  z.  Hence 
F  (x,  y,  z)  =  o  represents  a  surface  in  triplanar  coordinates. 

22.  If  the  equation  contains  only  two  variables  as  F  (x,y)  =  o  then 
it  represents  a  cylindrical  surface. 

For  F  (x,  y]  =  o  is  satisfied  by  certain  values  of  x  and  y  inde 
pendently  of  0,  and  x  and  y  are  no  longer  arbitrary  but  one  is  given 
in  terms  of  the  other  ;  to  each  pair  of  values  corresponds  a  point  in 
the  plane  xy,  and  the  locus  of  these  points  is  a  curve  in  that  plane. 
If  through  each  point  in  this  curve  we  draw  a  coordinate  parallel  to 
2,  every  point  in  that  coordinate  has  the  same  coordinates  x  andj/  as 
the  point  in  which  it  meets  the  plane  xy.  Hence  F  (x,y)  =  o  repre 
sents  a  surface  which  is  the  locus  of  straight  lines  drawn  through 
points  of  the  curve  F(.r,  v)  =  o  in  the  plane  xy  and  parallel  to  the 

12 


NOTES  ON  SOLID   GEOMETRY.  13 

axis  of  z.  This  locus  is  either  what  is  called  a  cylindrical  surface 
with  axis  parallel  to  z  or  a  plane  parallel  to  the  axis  of  z  according 
as  the  equation  F  (x,y)  —  o  in  the  plane  xy  represents  a  curve  or  a 
straight  line. 

For  example,  x1  4-  y*  —  r2=  o  in  rectangular  coordinates  is  a 
right  cylinder  wiih  circular  base  in  plane  xy  (since  jva+y  =  r*  is  a 
circle  in  plane  xy)  and  its  axis  coincident  with  the  axis  of  z. 

And  ax  +  by  —  c  =  o  is  a  plane  parallel  to  the  axis  of  g,  intersect 
ing  the  plane  xy  in  the  line  ax  +  by  =  c. 

Similarly  F  (x,  z)  =  o  represents  either  a  cylindrical  surface  with 
axis  parallel  tojy  or  a  plane  parallel  to_>'. 

F  (y,  z)  =  o  represents  either  a  cylindrical  surface  with  axis  parallel 
to  the  axis  of  x  or  a  plane  parallel  to  this  axis. 

23.  An  equation  containing  a  single  variable  represents  a  plane  or 
planes  parallel  to  one  of  the  coordinate  planes. 

Thus  x  =  a  represents  a  plane  parallel  to  the  planejyz. 

And  as  _/"(.#)  =  o  when  solved  will  give  a  determinate  number  of 
values  of  x,  as  x  =  a,  x  —  I,  x  =  c,  etc.,  so  it  represents  several 
planes  parallel  to  the  coordinate  planers. 

Thus  also  F(jy)  =  o  represents  a  number  of  planes  parallel  to  the 
plane  xz. 

And  F  (z)  =  o,  a  number  of  planes  parallel  to  xy. 

24.  Thus  we  see  that  in  all  cases  when  a  single  equation  is  inter 
preted  it  represents  a  surface  of  some  kind  or  other. 

The  apparent  exceptions  to  this  are  those  single  equations  which 
from  their  nature  can  only  be  satisfied  when  several  equations  which 
must  exist  simultaneously  are  satisfied.  As  for  example 

(x  —  of  +  (y  —  b)*  +  (z  —  c)z  =  o.  This  equation  can  only  be 
satisfied  when  (x  —  a)*  =  o,  (y  —  b}*  =  o,  (z  —  c}z  =  o,  or  x  =  a, 
y  —  b,  z  =  c. 

Now  these  represent  three  planes,  but  being  simultaneous  they 
represent  the  point  a,  b,  c. 

So  also  (x  —  a)z  -f  ( y  —  b}*  =  o  is  only  satisfied  by  x  =  a,  y  =  b, 
and  hence  though  x  =  a  is  a  plane,  and  y  =  b  is  a  plane,  the  two 
together  must  represent  a  line  common  to  both  of  these  planes,  that 
is  their  line  of  interseciion,  which  must  be  parallel  to  z. 

25.  In  general  two  simultaneous  equations  as 

f(*,y,  *)  =  °  F  (x,y,  z)  =o 


14  NOTES  ON  SOLID   GEOMETRY. 

represent  a  curve   or  curves,   the   intersections  of  the   two  surfaces 
represented  by  the  two  equations. 

Thus    _  7  r  taken  simultaneously  we  have  seen  represent  a  straight 

line  parallel  to  the  axis  of  z,  the  intersection  of  these  two  planes. 

F  (x)  =  o  ) 

p  )   \  _  0  f  represent  a   number   of  straight  lines  parallel  to   the 

axis  of  z,  the  intersections  of  the  several  planes  parallel  respectively  to 
the  planes  yz  and  xz. 

F  (x)  =  o  ) 

p/j)  —of  rePresent  a  number  of  straight  lines  parallel  to  the  axis 

of  y,  e;c. 

f.  represent  the  curves  of  intersection  of  the  two  cylin- 
P  (.v,  z)  =  o  j  * 

ders  F  (x,  y)  =  o  and  F  (x,  z)  =  o,  e'c.,  etc. 
26.  Three  simultaneous  equations 

F  (x,  y,  z}=o\  F  (x,  j>)  =  o\ 

as   f(x,  y,  z)  =  o  I       or    F  (A-,  z)  =  o  I  etc., 

represent  points  in  space  or  the  intersections  of  the  lines  of  intersec 
tion  of  the  surfaces. 
The  simplest  case  is, 


y  —  ^  >  representing  the  point  (a,  b,  c], 

z  =  c] 
So  also 


2z  \ 

x  +  y  =  2z    v.  represent  points  which    can  be  found  by 


solving  the  three  equations  which  themselves  represent  different  sur 
faces. 

Interpretation  of  Polar  Equations. 

27.  i°.  r  =  a  represents  a  sphere  having  the  pole  for  its  centre. 
Hence  the  equation  ¥  (r)  =  o  which  gives  values  for  r  as  r  —  a, 
r  =  b,  r  —  c,  etc.,  represents  a  series  of  concentric  spheres  about  the 
pole  as  centre. 


NOTES   ON  SOLID   GEOMETRY.  jcj 

2°.  6  =  a  represents  a  cone  of  revolution  about  the  axis  of  z  with 
its  vertex  at  the  origin  of  which  the  vertical  angle  is  equal  to  2a. 
Hence  the  equation  F  (6)  =  o  giving  values  6  =  a,  6  =  ft,  etc., 
represents  a  series  of  cones  about  the  axis  of  z  having  the  origin  for 
a  common  vertex. 

3°.  cp  =  (3  represents  a  plane  containing  the  axis  of  z  whose  line 
of  intersection  with  the  plane  xy  makes  an  angle  /3  with  the  axis  of 
x.  Hence  the  equation  F  (cp)  —  o  which  gives  values  q)  =  (3,  cp 
=  ft',  etc.,  represents  several  planes  containing  the  axis  of  z  inclined 
to  the  plane  zOx  at  angles  ft,  ft',  etc. 

4°.  If  the  equation  involve  only  rand  6  as  F  (r,  6)  =  o,  since 
F  (r,  6)  =  o  gives  the  same  relation  between  r  and  6  for  any  value 
of  cp,  ii  gives  the  same  curve  in  any  one  of  the  planes  determined  by 
assigning  values  to  (p.  Hence  it  represents  a  surface  of  revolution 
traced  by  this  curve  revolving  about  the  axis  of  z. 

Example,  r  =  a  cos  6  is  the  equation  of  a  circle  in  the  plane 
xz,  or  in  any  plane  containing  the  axis  of  z.  Hence  r  —  a  cos  6 
represen  s  a  sphere  described  by  revolving  this  circle  about  the  axis 
of*. 

5°.  If  the  equation  be  F(<p,  0)  —  o  for  every  value  of  cp  there 
are  one  or  more  values  of  6  corresponding  to  which  lines  through 
the  po'e  may  be  drawn,  and  as  cp  changes  or  the  plane  fixed  by  it 
containing  Oz  revolves,  these  lines  take  new  positions  in  each 
new  position  of  the  plane,  and  thus  generate  conical  surfaces 
(a  conical  surface  being  any  surface  generated  by  a  straight  line 
moving  in  any  manner  about  a  fixed  straight  line  which  it  inter 
sects.  ) 

6°.  If  the  equation  be  F(r,  (p)  =  o,  for  every  value  of  (p  there  are 
one  or  more  values  of  r,  thus  giving  several  concentric  circles  about 
the  pole  in  the  plane  determined  by  the  assigned  value  of  cp.  As  (p 
changes,  or  the  plane  through  Oz  revolves  these  values  of  r  change, 
and  the  concentric  circles  vary  in  magnitude.  The  equation  thus 
represents  a  surface  generated  by  circles  having  their  centres  at  the 
pole,  which  vary  in  magnitude  as  their  planes  revolve  about  the  axis 
of  z  which  they  all  contain. 

7°.  If  the  equation  be  F  (r,  6,  cp)  =  o,  it  represents  a  surface  in 
general.  For  if  we  assign  a  value  to  <p  as  cp  =  ft,  then  F  (r,  8,  ft) 
=  o  will  represent  a  curve  in  the  plane  (p  =  ft.  And  as  cp  changes 
or  the  plane  revolves  about  Oz  this  curve  changes,  and  the  equation 
will  represent  the  surface  containing  all  these  curves. 


!6  NOTES  ON  SOLID   GEOMETRY. 

28.  Two  simultaneous  equations  in  polar  coordinates  represent  a 
line,  or  lines — the  intersections  of  two  surfaces.  And  three  simulta 
neous  equations  represent  a  point  or  points — the  intersections  of  three 
surfaces. 

Thus 

r  =  a  j 

6  —  a  v  taken  simultaneously  represent  points  determined 

9  =  ft) 

by  the  intersection  of  a  sphere,  cone  and  plane. 


CHAPTER  III. 

EQUATION   OF   A   PLANE. 

COORDINATES   OBLIQUE   OR   RECTANGULAR. 

29.  To  find  equation  of  a  plane  in  terms  of  the  perpendicular  from  the 
origin  and  its  direction  cosines. 

Let  OD  —p  be  the  perpendicular  from  the  origin  on  the  plane, 
and  let  it  make  with  the  axes  O,r,  Oy  and  Qz  the  angles  a,  ft  and  y 
respectively.  Let  OP  be  the  radius  vector  of  any  point  P  of  the 
plane  ;  OM,  MN  and  NP  the  coordinates  of  P.  (Fig.  n.) 

The  projection  of  OM  +  MN  +  NP  on  OD  is  equal  to  the  pro 
jection  of  OP  on  OD. 

The  projection  of  OP  on  OD  is  OD  itself,  and  the  projection  of 
OM  +  MN  +  NP  on  OD  is  x  cos  a  +y  cos  ft  +  z  cos  y. 

Hence  we  have         x  cos  a  +y  cos  ft  +  z  cos  y  —p.     (12) 

30.  To  find  the  equation  of  a  plane  in  terms  of  its  intercepts  on  the 
coordinate  axes  (coordinates  oblique  or  rectangular). 

Let  the  intercepts  be  OA  —  a,  OB  =  b,  OC  =  c.  The  equation 
(12)  may  be  written 

x  y  z 

p  sec  a      p  sec  ft       p  sec  y 

But  since  ODA,  ODB  and  ODC  are  right-angled  triangles,  we  have 
p  sec  a  =  OA  —  a,  p  sec  ft  =  OB  =  b,  p  sec  y  =  OC  =  c. 
Therefore  the  equation  becomes 

X        V         Z 


the  equation  of  the  plane  in  terms  of  its  intercepts. 


1 8  NOTES   ON   SOLID   GEOMETRY. 

31.  Any  equation  Ax  +  By  +  Cz  =  D  (14)  of  the  first  degree  in  x, 
y  and  z  is  the  equation  of  a  plane* 

For  we  may  write  (14) 

x      j^     _z_ 

Jl    J>/ _D/ 

X     "B"     ~C~ 

D  D  D 

And  putting -^=  a, -g- =  ^ -£-=  A  We    have   the   form    (13). 

Hence  (14)  is  the  equation  of  a  plane  in  oblique  or  rectangular 
coordinates. 

Hence  to  find  the  intercepts  of  a  plane  given  by  its  equation  on 
the  coordinate  axes,  we  either  put  it  in  the  form  (13)  or  simply 
raakej>  =  o  and  z  =  o  to  find  intercept  on  x  ;  z  =  o  and  x  =  o  to 
find  intercept  onj/ ;  x=  o  andjy  —  o  to  find  intercept  on  z. 

"Example.     Find  the  intercepts  of  the  plane  2x  +  $y  —  52  =  60. 

32.  It  is  useful  often  to  reduce  the  equation  AJC  +  By  +  Cz  =  D 
to  the  form  x  cos  a  +  y  cos  ft  +  z  cos  y  =/  in  rectangular  coordi 
nates.      We  derive  a  rule  for  this. 

Since  both  of  these  equations  are  to  represent  the  same  plane,  we 
have 


cos  a  _  cos  (3  _  cos  y  _  p  _  Vcos2  a.  +  cos2  (3  +  cos2  y 
A  B  C        D     yT7       A2  +  B2  +  C2"" 


Hence  cos  <*  = 


+  B2  +  C2  -v/A2  +  B2  +  C 


+  B'  +  C2  A/A2  +  B2+C! 

=  -vWWc*    (I5) 

it  is  in  the  perpendicular  form  (12). 


NOTES  ON  SOLID   GEOMETRY.  ig 

Hence  the  Rule:  If  we  divide  each  term  of  the  equation  Ax  +  By 
=  D,  by  the  square  root  of  the  sum  of  the  squares  of  the  coefficients 
ofx,  y  and  z,  the  new  coefficients  will  be  the  direction  cosines  of  the  per 
pendicular  to  the  plane  from  the  origin,  and  the  absolute  term  will  be  the 
length  of  this  perpendicular.  Give  the  radical  the  sign  of  D. 

Example.      Find  the  direction  cosines  of  the  plane  2x  +  $y  —  40 
=  6  and  the  length  of  the  perpendicular  from  the  origin. 

Result. 

2  2  3  —  4 

cos  a  =  ===  ——7=,  cos  ft  =  —T=,  cos  y  =——, 

A/4  +  9  +  16       v  29  V  29  V  29 


P  = 


V  29 

33.   To  find  the  angle  between  two  planes   (coordinates    rectangu 
lar). 

If  the  planes  are  in  the  form 

x  cos  a  +  y  cos  ft  +  z  cos  y  =  p 
x  cos  a'  +  y  cos  ft'  +  z  cos  y'  —  p', 

then  since  this  angle  is  equal  to  the  angle  of  two  perpendiculars  from 
origin  on  the  planes  the  cosine  will  be  (Art.  15)  cos  V=  cos  a  cos  a' 
+  cos  /3  cos  /3'  -f  cos  y  cos  y1. 
If  they  are  in  the  form 

A.v  +  By  +  Cs  =  D 
A'.v  +  B>  +  C'z  =  D'. 

Then  cos  a  = — •==.  r,  cos  /?  = 

B2  4-  C2  A/A2+  B2  +  C2 

C 


cos     = 


A'  al  B' 

COS  flf    =  —  ~,   COS  P    =  — 


A/A'2  +  B/f+ 


And    cos  V— 


cos  y  = = 

AA'  +  BB'  +  CC' 


AA2  +  B2  +  C  VA/f  +  B'2  +  C'2 


2Q  NOTES   ON  SOLID  GEOMETRY. 

From  this 
.  Q      _  (A2-f  B2+  C2)  (A'2  +  B'2  +  C'2)  -(AA'  +  BB'  +  CC')2 

(A2  +  B*  +  C2)(A'2+B'2+  C"2) 

d  .  v  -  (AB'-A'B)2+(AC'-A'C)2+(BC'-B'C)2 
(A2  +  B'  +  C')  (A'8+  B'a  +  C'2) 

Cor.  i.  If  the  planes  are  perpendicular  to  each  other,  then  cos  V=o. 
.'.  AA'  +  BB'+  CC'  =  o  (18)  is  the  condition  of  perpendicularity  of  the 
planes. 

Cor.  2.   If  the  planes  are  parallel  sin  V  =  o.     Hence 

(AB'  -  A'B)2  +  (AC'  -  A'C)2  +  (BC'  -  B'C)2  =  o 
or  AB'  -  A'B  =  o    AC'  -  A'C  =  o     BC'  -  B'C  =  o 

ABC 
A7==B7=C 

or  the  condition  that  the  two  planes  shall  be  parallel,  is  that  the  coefficients 
ofx,y  and  z  in  the  two  equations  shall  be  proportional. 
Ex.  i.  Find  the  angle  between  the  planes 

x  +  2y  +  32  =  5  and  $x  —  4y  4-  z  =  10. 

2.  Show  that  the  planes 

x  +  3y  —  5Z  —  20  anc*  2X  +  V  +  z  —  I0   are  perpen 
dicular  to  each  other. 

3.  Write  the  equation  representing  planes  parallel  to  the  plane  3* 
+  2y  —  6z  =  ii. 

34.    To  find  the  expression  for  the  distance  from  a  point  P  (x'y'z')  to  a 
plane  (coordinates  rectangular). 

i°.   Let  the  equation  of  the  plane  be  of  the  form 

x  cos  a  +  y  cos  ft  +  z  cos  y  =  /  when  p  —  OD. 

Pass  a  plane  through  P  parallel  to  the  given  plane,  and  produce 
OD  to  meet  it  in  D'. 

The  equation  of  this  plane  will  be 

x'  cos  a  +y  cos  ft  +  z  cos  y  =  p'  when  OD'  =/'. 

Now  let  PM  be  the  perpendicular  from  P  on  the  given  plane. 
Then  PM  =  OD'  -  OD  =  p1  -  p. 


NOTES   ON  SOLID    GEOMETRY.  21 

Hence  PM  =  x'  cos  a  4-  y  cos  /3  +  z  cos  y  —  p. 

And  ±  (x1  cos  a  +  y'  cos  /?  +  0'  cos  y  —  p)  (20)  is  the  expression 
for  the  perpendicular  from  the  point  x'y'z'  on  the  plane  x  cos  or  +  _y 
cos  ft  +  0  cos  y  =p,  the  sign  being  +  or  —  according  as  P  is  or  is 
not  on  the  side  of  the  plane  remote  from  the  origin. 

2°.  Let  the  plane  be  in  the  form  Kx  +  By  +  Cz  =  D. 

A 

Then  cos  a  =—  -  etc.,  etc.    (15)  Art.  32. 

V  A2  +  Bs  +  C* 

Hence  the  expression 

±(y  cos  «  -FJF'  cos  /3  +  z'  cos  y  —  p]  becomes 
-D 


Ex.   Find  the  length  of  perpendicular  from  the  point  (3,  2,  i)  on 

the  plane 

3Jt-  +  4_y  —  6z  =  24. 

9  4-  8  -  6  —  24        —13 
Result.  p  —  — 


-  - 

A/9+  16+  36 

35.  The  equation  of  the  plane  in  the  form  x  cos  a  +  y  cos  (3  4-  0 
cos  y  =  p  may  be  used  to  demonstrate  the  following  theorem  in 
projec  ions. 

7%£  volume  of  the  tetrahedron  which  has  the  origin  for  its  vertex  and 
the  triangle  ABCy2?r  its  base  is  equal  to  the  three  pyramids  which  have  any 
point  (x,  y,  z)  in  the  plane  ABC  for  their  common  vertex  and  for  bases 
the  projections  of  the  area  ABC  on  the  three  rectangular  coordinate  planes 
respectively. 

For  let  A  be  the  area  of  the  triangle  ABC  and 

x  cos  a  +y  cos  /3  +  z  cos  y  =  p 
the  equation  of  its  plane. 

Multiply  this  equation  by  A. 

Then         A  cos  a  .  x  -f  A  cos  /3  .  y  +  A  cos  y  .  z  =  Ap 
or  -JA  cos  a  .  x  +  ^A  cos  ft  .y  +  -JA  cos  y.  s  — 


But  A  cos  a,  A  cos  /3,  A  cos  y,  are  the  projections  of  A  on  the 
planes  yz,  xz,  and  xy  respectively,  and  x,  y  and  z  are  the  altitudes  of 
the  tetrahedrons  which  have  these  projections  as  bases  and  the  point 
(x,  y,  z)  as  common  vertex,  and  -JA/>  is  the  volume  V  of  the  pyramid 


22  NOTES  ON  SOLID   GEOMETRY. 

which  has  the  origin  for  vertex  and  A  for  base.  Hence  the  theorem 
is  true. 

Calling  these  projections  Ax,  Ay,  and  Ax,  we  may  write  the  equa 
tion  of  the  plane  A^v  +  Ayy  +  A^  =  3V.  (22) 

36.  To  find  the  polar  equation  of  a  plane . 

Let  OP  =  r,  POS  =  6,  P'OM  =  cp  be  the  polar  coordinates  of  a 
point  P  of  the  plane.  (Fig.  12.) 

Let  OD  =  a  =  perpendicular  on  plane  ;  angle  DOS  —  a,  D'OM 
=  jff,  and  POD  =  GO. 

Then  yyp-  =  cos  POD  =  cos  GO,  or  —  =  cos  GO.     Now  in  order  to 

express  GJ  in  polar  coordinates  conceive  a  sphere  about  O  as  centre 
with  OP  =  r  as  radius.  Prolong  OD  to  D"  on  the  sphere.  Draw 
the  arcs  of  great  circles  SPP',  SD"D',  MP'D'  and  D"P. 

The  triangle  SD"P  has  for  its  sides  SD"  =  a,  SP  =  6,  D"P  =  GJ 
and  angle  D"SP  =  D'OP'  =  /3  —  <p.  But 

cos  D"P  =  cos  SD"  cos  SP  +  sin  SD"  sin  SP  cos  D"SP. 

Or 

cos  GO  =  cos  a  cos  6  +  sin  a  sin  6  cos  (fi  —  cp). 

Therefore  —  =  cos  a  cos  6  +  sin  a  sin  6  cos  (f3  —  cp)  (23)  is  the 
polar  equation  of  the  plane. 

37.  The  general  equation  of  the  plane  Ax  +  By  4-  Cz  =  D  may 
be  reduced  to  the  form 

A'x  4   Bfy  +  C'2  =  i    (24)  by  dividing  by  the  absolute  term  D. 

And  also  to  the  form 

z  —  mx  +  ny+  c  (25)  by  dividing  by  C  —  transposing  and  putting 

=  m,  —  —  =  n  and  -^  =  c.    These  two  forms  are  very  useful  in 

V^/  V^  v_/ 

the  solution  of  problems  and  in  finding  the  equations  of  the  plane 
under  given  conditions. 

Plane  tinder  Given  Conditions. 

38.  i°.   The  equation  of  a  plane  through  the  origin  will  be  of  the 
general  form  Ax  +  By  +  Cz  =  o,  for  ihe  equation  must  be  satisfied 
by  ^  =  0,^  =  0  and  2  =  0. 

2°.   The  equation  of  a  plane  which  contains  the  axis  of  z  is  of  the 


NOTES 


SOLID  GEOMETRY, 


form  Ax  +  By  =  o  ;  a  plane  containing  the  axis  of  y  is  Ax  +  Gar 
—  o  ;  one  containing  the  axis  of  x  is  By  +  Cz  =  o. 

3°.  The  equation  of  a  plane  parallel  to  the  axis  of  z  is  Ax  +  By 
=  D;  of  one  parallel  to  the  axis  ofjy  is  AJI*  +  Cz  =  D  ;  one  parallel 
to  the  axis  of  x  is  By  +  Cz  =  D. 

4°.  The  equation  of  a  plane  parallel  to  the  plane  j/s  is  A^t:  =  D  ; 
parallel  to  ^0  is  By  =  D  ;  parallel  to  xy  is  Cs  =  D. 

These  equations  we  have  had  already  in  the  forms  x=  ±a,  y—  ±3, 


39.  To  find  the  equation  of  a  plane  containing  a  given  point  (a,  b,  c) 
and  parallel  to  a  given  plane  Ax  -f  By  +  Cz  =  D.  (  i  ) 

First,  since  the  required  plane  is  to  be  parallel  to  (i  )  it  may  be  writ 
ten  A*  +  Bj/  +  G&=D'  (2)  when  D'  is  undetermined.  Secondly,  the 
coordinates  (a,  b,  c)  must  satisfy  (2).  Therefore  Aa  +  ~B&  +  Cc  =D'. 
Hence  by  subtraction  we  eliminate  D'  and  obtain 


Ax  +  By  +  C0  =  Aa  +  B3  +  O     (26) 
the  required  equation. 

Example.    Find  the  equation  of  the  plane  passing  through  the 
point  (i,  2,  4)  parallel  to  the  plane  2x  +  4y  —  32  =  6. 

40.   To  find  the  equation  of  a  plane  passing  through  three  given  points 
(x',  y',  z'),  (x",  y",  z")  and  (x'",  y'",  z'"). 

Let  the  equation  of  the  plane  be  of  the  form  Ax  +  By  +  Cz  =  i, 
A,  B  and  C  to  be  determined  by  the  given  conditions. 

Since  the  plane  is  to  contain  each  of  the  points,  we  must  have 

Ax'  +  B 


Hence 


I,/", 


B  = 


X,    I,   ^ 

JP",  i,  z 
x'",  i,  2 


C  = 


Substituting   these    values   in   the    equation    Ax  +  By  +  Cz  =  i 


NOTES  ON  SOLID   GEOMETRY. 


we  have 


I,/', 


#,  I,  * 

*,    I,    * 


x  +  x"r,  i, 


(27) 


But  from  plane  coordinate  geometry  the  coefficients  of  x,  y  and  z 
in  these  equations  are  the  double  areas  of  triangles  in  the  planes  jy0, 
xz  and  xy  respectively.  Moreover  these  triangles  are  the  projections 
of  the  triangle  of  the  three  given  points,  on  these  planes.  Hence 
comparing  this  equation  wi  h  the  equation  (22) 


we  see  that 


x  ,y  ,ss 

x">y", 2 


,z  =  3V 


=  6V.    That  is  =  6  times  the  volume  of  the 


pyramid  which  has  the  origin  for  vertex  and  the  triangle  of  the  three 
given  points  for  base.  This  equation  fully  written  out  is 

^(yv--yv')+^''(yv-y3''')+y''(jV'-yv)-6v,   (28) 

41.  To  find  the  equation  of  the  planes  which  contain  the  line  of  intersec 
tion  of  the  two  planes  Ax  +  By  +  Cz  =  D  and  Ax'  4-  By'+  Cz'  =  D'. 

ThisequationisA#  +  By  +  Gs— D  +  K(A'jc  +  B>  +  C'0-D/)=o(29) 

when  K  is  arbitrary.  For  this  represents  a  plane  when  K  takes  a 
particular  value  and  it  is  satisfied  when  A^  +  By  +  C-s:  —  D  =  o  and 
K'x  4-  B^  +  C'z  —  D'  =  o  are  satisfied  simultaneously.  Hence  it 
is  a  plane  containing  their  line  of  intersection.  Hence  as  K  is  arbi 
trary  it  (24)  represents  the  planes  containing  the  line  of  intersection 
of  the  two  given  planes. 

42.  When    the  identity  KU  +  K^Uj  +  KaUs  =  o    (30)    exists  between 
the   equations   U=o,  U2  —  o,  U3=  o    of  three  planes,  then   these  planes 
intersect   each    other  in  one  and  the  same  straight  line.       This   is  an 
easy  corollary  of  Article  41.     Also  when  the  equation  of  the  first 
degree  in  x,  y  and   z  contains  a  single  arbitrary  constant  all   the 
planes  which  it  expresses  by  assigning  particular  values  to  this  con 
stant  intersect  each  other  in  one  and  the  same  straight  line.     This 
line   of  intersection  may  be  at  infinity  and  then  the  planes  are  all 
parallel. 

Example  I.  The  planes  represented  by  the  equation  6x+T&y  +  2z 
=.  3  (M  arbitrary)  all  contain  the  line  of  intersection  of  the  two 
planes  6x+  22  —  3  =  0  and  y  —  o. 


NOTES  ON  SOLID   GEOMETRY.  25 

Example  2.  The    planes   represented    by   2x  +  ^y  —  42  =  n    (n 
arbitrary)  are  parallel. 

Example.     The  planes  $x  +  \v  +  60  =  2  } 

X  +   2V  +  3£  —  I   ?• 
4.v  +  4T  +  &z  =  2  J 

intersect  in  one  and  the  same  straight  line  because 

—  2   =  o 


is  an  identity. 

43.  Wfo«  between  the  equations  of  four  planes  in  any  form  U  —  o,  U, 
=  o,  U2  —  o,  U3  —  o  the  identity 

KU  +  K1U1  +  K2Ua-hK3U3  =  o  (31)  ^/j/j,  then  these  four  planes 
intersect  each  other  in  one  and  the  sam*  point.  For  then  any  coor 
dinates  which  satisfy  the  first  three  U  =  o,  U,  =  o  and  U2  =  o  will 
satisfy  the  fourth  U3  =  o. 

44.  Example  i.    Find  the  equation  of  the  plane  passing  through 
the  origin  and  containing  the  line  of  intersection  of  the  two  planes 
Ax  +  By  +  C0  =  i  and  A'x  +  B>  +  C'z  =  i. 

First  we  have  Ax  +  By  +  Cz  —  i  +  K  (A'.r  +  B>  +  C'z  —  1)=  o 
for  all  the  planes  containing  the  line  of  intersection  of  the  two  given 
planes.  But  as  the  required  plane  must  contain  the  origin,  the 
equation  must  be  satisfied  by  (o,  o,  o).  Hence  we  have  —  i—  K=o. 
.-.  K  =  -  i. 

The  required  equation  is  therefore 

Ax  +  By  -f  Cz  —  i  —  (A'x  +  B>  +  C'z  —  i)  =  o 
or  (A  -  A')  x  4-  (B  -  B')^  +  (C  -  C')  z  =  o. 

Ex.  2.  On  the  three  axes  of  x,  y  and  z  take  OA  =  a,  OB  —  b, 
OC  =  c  and  construct  on  these  a  parallelopipedon  having  MP  as  the 
edge  opposite  parallel  to  OC,  and  AR  in  the  plane  xz  the  edge 
opposite  and  parallel  to  BN. 

Find  the  equation  to  the  plane  containing  the  three  points  M,  N 
and  R. 

\?  V 

Now  NR  is  the  line  of  intersection  of  the  two  planes  —  +-7-  =  i  and 

a        b 

-=  i.      Hence  the  plane  containing  this  line  must  be  of  the  form 


26  A'OTES  ON  SOLID   GEOMETRY. 

OC         V 


"*~  ^  I  --  l  )  —  °-     To   determine   K  \ve  impose  the 
condition  that  this  plane  shall  pass  through  the  po'nt  M  (a,  b,  o). 

Hence  we  have  —  +  -r  —  i  +  K  (  —  i  )  =  o.     .  •.  K  =  i 
at  \c     J 

Therefore  the  required  plane  is 

x       v  z  x       v      z 

--t-V  —  H  ----  i  —  o  or  --  (-^-_j-__  —  2. 
a       b  c  a       b       c 

Ex.  3.  Find  in  like  manner  the  equation  of  the  plane  containing 
the  points  P,  B,  and  C,  in  the  same  figure. 

Result, 

45.    If  two  given  planes  be  in  the  normal  form  as 
x  cos  a  +>'  cos  ft  +  z  cos  y=p  and.v  cos  a'  +y  cos  /3'  +  zcos  y'  =  p'. 
The  plane  containing  their  line  of  intersection  is 

x  cos  a+y  cos  /3  +  z  cos  y  —  p-\-K  (x  cos  a  +y  cos  ft'  +  z  cos  y 

-/)=o 

And  if  K  =  ±  i  the  equation  becomes 

x  cos  a  -\-y  cos  /3  +  z  cos  y  —  p  ±  (x  cos  a  +  y  cos  ft'  +  z  cos  ;/  ' 

-/>')=  o 

which  represents  ihe  two  plane  bisectors  of  the  supplementary  angles 
made  by  the  given  planes. 

That  is  to  find  the  equations  to  the  plane  bisectors  of  the  supplementary 
angles  made  by  two  given  planes,  put  their  equations  in  the  normal  form 
and  then  add  and  subtract  them. 

Example.  Find  the  two  planes  which  bisect  the  supplementary 
angles  made  by  the  planes  2.v  +  3.y  Vz  =  5  and  3^  +  4^—23  =--  4- 


Result,  .  ±..=  0. 

A/14  V29 

Remark.  If  we  place  A  =  x  cos  (Y  +  y  crs  ft  +  z  cos  y  —  p  and 
A'  =  x  cos  a'  +  v  cos  /?'+  0  cos  y'  —  />'. 

Then  A'—  A  =  o  is  the  plane  bisector  of  one  of  the  angles  be 
tween  the  planes  A  and  A'  and  A  -f  A'  =  o  is  that  of  the  supple 
mentary  angle. 

46.  The  three  planes  which  bisect  the  diedral  angles  of  a  triedral  have. 
a  common  line  of  intersection.  Let  A  =  o,  A'=  o  and  A"  =  o  be  three 


NOTES   ON  SOLID   GEOMETRY. 


planes  in  the  normal  form,  and  let  the  origin  be  within  the  triedral angle 
formed  by  the  three  of  which  P  their  point  of  intersection  is  the  vertex. 

Then  the  plane  bisectors  of  the  angles  made  by  these  planes  is 
A  —  A'  —  o,  A"—  A  —  o,  A'—  A"  =  o.  And  as  these  when  added 
together  vanish  simultaneously,  it  follows  that  these  three  planes 
have  a  common  line  of  intersection. 

We  can  give  this  theorem  another  form  by  conceiving  a  sphere  to  be 
described  about  the  vertex  of  the  triangular  pyramid  as  a  centre.  The 
three  planes  A=  o,  A'=  o,  A"=o  cut  the  surface  of  the  sphere  in 
arcs  of  great  circles  which  form  a  spherical  triangle  and  the  three 
planes  A  —  A'=o,  A"—  A  =  o  and  A' —  A"  =  o  cut  the  sphere  in 
three  arcs  of  great  circles  which  bisect  the  angles  of  this  spherical 
triangle  and  their  common  line  of  intersection  pierces  the  sphere  in 
the  common  intersection  of  these  arcs.  Hence  the  above  demon 
strates  the  following  theorem,  namely,  The  arcs  of  great  circles  which 
bisect  the  angles  of  a  spherical  triangle  cut  each  other  in  the  same  point 
(the  pole  of  the  inscribed  circle  of  the  triangle). 

47.  To  find  the  point  of  intersection  of  the  planes  AJC  +  By  +  Cs 
=  D,  A'x  +  B>  +  C'z  =  D',  A",r  +  B'>  +  C"z  =  D". 

We  have  by  elimination 


D,  B,  C 
D',  B',  C' 
D".  B",  C" 


A,     D,    C 

A',   D',   C' 
A",  D",  C" 


A,    B,     D 
A',   B',    D' 

A",  B",  D" 


(32) 


A,     B,    C  [A,     B,    C  A,     B,     C 

A',    B',  C'  A',    B',  C'  A',    B',   C' 

A",  B",  C"  A",  B",  C"  A",  B",  C" 

Hence  the  condition  that  one  of  these  shall  be  parallel  to  the  line 
of  intersection  of  the  other  two,  or  that  the  planes  shall  not  meet  in 
a  point,  is 

A,     B,     C 

A',    B',    C' 

A",  B",  C"    =  o,  that  is 

A(B'C"-B"C')  +  A'(B"C  -  BC")  +  A"(BC'  -  B'C)  =  o. 
47.   The  condition  that  four  planes 

Ax    -fBy    +Cz     +D    =  o  1 

A'x  +  B>   +  Cz    +  D'   =o 

A"x  +B'>  +C"z  +D"  =o 

A'"x+B'"y  +  C'"z  +  ~D'"  =  o      shall  meet  in  a  point  is 


28 


NOTES   ON   SOLID   GEOMETRY. 


A,  B,  C,     D 

A',  B',  C',    D' 

A",  B",  C",  D" 

| A'",  B'",  C",  D'" 


=  o.    (33) 


49.  We  have  seen  that  the  equations  of  two  planes  Ajr 
—  D  =  o  and  A.'x+  Bjy  +  Cz  —  D'  =  o  added  together  one  or  both 
of  them  multiplied  by  any  number  give  the  equation  of  a  plane 
which  contains  the  line  of  intersection  of  the  two  given  planes.  If 
we  combine  these  two  equations  so  as  to  eliminate  x  we  shall  obtain 
a  plane  parallel  to  the  axis  of  A\  containing  this  line  of  intersection. 
If  we  eliminate  y  we  obtain  a  plane  parallel  to  the  axis  of  y  contain 
ing  the  same  line  ;  and  finally  if  we  eliminate  z  we  obtain  a  plane 
parallel  to  the  axis  of  z  containing  the  same  line. 


CHAPTER    IV. 

THE    STRAIGHT    LINE. 

50.    The  equations  of  any  two  planes  taken  simultaneously  represent 
their  line  of  intersection. 

rePresent  a  straiSht  lme  the  c°- 


ordinates  of  every  point  of  which  will  satisfy  the  two  equations. 

If  we  eliminate  alternately  x  and  y  between  these  equations  we 
obtain  equations  of  the  form 

x  —  m        p\        ,     x         ^    planes    parallel    respectively    to    the 

y  ~  nz   +  q  \ 

axes  Oy  and  Ox  which  represent  the  same  straight  line  as  equations 
(35).  These  non-symmetrical  forms  (35)  are  very  useful.  The 
planes  x  •=.  mz  +  a,  y  =  nz  +  b  are  called  the  projecting  planes  of  the 
line  on  the  planes  of  xz  and  jar,  and  these  equations  are  also  the 
equations  of  the  projections  of  the  line  on  those  planes  respectively. 

If  we  eliminate  z  we  get  -  —  ~=  --  —  or  y  =  —  x  +  g  --  p  the  equa- 
n  m        J       m  m 

tion  of  the  projection  of  the  line  on  the  coordinate  plane  xy. 

The  equations  (35)  of  the  straight  line  contain  four  arbitrary  con 
stants,  m,  n,  p,  q,  to  which  we  can  give  proper  significance  by  com 
paring  these  equations  with  the  equation  y  —  mx  +  b  in  plane  coor 
dinate  geometry. 

The  equations  (35)  may  be  thrown  in  the  form 

^_^=z_^  =  , 

m  n  i 

which  gives  us  an  easy  choice  of  fixing  the  line  by  the  equations 
of  any  two  of  its  projecting  planes. 

51.  To  find  the  equations  of  a  straight  line  in  terms  of  its  direction  cosines 
and  the  coordinates  a,  b,  c  of  a  point  on  the  line  :  (axis  rectangular.} 

29 


30  ATOTES   ON   SOLID    GEOMETRY. 

Let  a,  ft,  y  be  the  angles  made  by  the  line  with  the  coordinate  axes 
respectively.  Let  /  be  the  portion  of  the  line  between  any  point 
(x,  y,  z)  on  the  line  and  the  point  (a,  b,  c).  Then  /  cos  a  =  a-— a  ; 
/  cos  ft  —y—b  ;  /  cos  y—  z—c  •  and  eliminating  /  we  have 

(37) 


cos  ex      cos  ft       cosy 

This  form  (37)  of  the  equation  of  a  straight  line  is  symmetrical 
and  is  therefore  very  useful.  It  contains  six  constants  but  in  reality 
only  four  independent  constants,  since  the  relation  cos2  o'  +  cos2  ft 
4- cos2  y  =  i  holds,  and  of  the  three  a,  b,  c  one  of  them  may  be 
assumed  at  will,  leaving  only  two  independent. 

We  have  seen  that  the  equation  (35)  may  be  thrown  into  the  form 
(37).  So  also  (37)  may  be  thrown  into  the  form  (35)  by  finding 
from  them  expressions  forjy  and  x  in  terms  of  z. 

52.  To  find  the  direction  cosines  of  any  straight  line  given  by  its  equations. 

If  the  equations  be  in  the  form 

x—  a        v—bz—c       T      HT         i   XT  •        i 

— = —  =  — — —  r^  — — — .      L,    M   and  N  are  proportional   to  the 

direction  cosines  of  the  line. 
So  that  we  have 


cos  a  _  cos  ft  _  cos  y  __  -\Xcos-  a  +  cos2  ft  +  cos2  y 
L        ~M~~        N  ~ 


Hence 

N 


Hence  to  find  the  direction  cosines  of  any  straight  line 

A.r  +  By  +  Cz  =  D 
MX  4-  B>  +  C'z  =  D' 

we  throw  the  equations  into  the  form 

.v—a  _  y—b  _  z— c 
L        "IT"    '~N~ 

by  eliminating  y  and  x,  and.  then  write  out  the  direction  cosines  as  above 
equal  to  each  denominator  divided  by  the  square  root  of  the  sum  of  the 
squares  of  all  three. 


NOTES   ON  SOLID   GEOMETRY. 

Thus  to  find  the  direction  cosines  of  the  line  1 

y  =  nz  +  q  \ 

X  —  p         V  —  Q         Z 

" 


we  write  it 

m  n 

Hence 

m 


•  cos   =^+*+r.  cos       *> 

(39) 


Ex.   i.    Find  the  direction  cosines  of  the  lines 
x  —  5        y  ~  2       z  + 


,,  .  _  =  24  .  v 

.,T^  3^-1  j  \  >  >     3x—4y+2z  -  10  U;' 

53.    To  find  the  cosine  of  the  angle  between  two  lines  given  by  the  equa 
tions 

x—a       y—b       z— c         ,  x—  a        v—b'       z—c 
and  — ri—  =: 


L  M  N  L'  M'          N' 

We  have  shown  (Art.  15) 

cos  V  =  cos  a  cos  a'-f  cos  /3  cos  /?'+  cos  ^  cos  y. 

LL'  +  MM'  +  NN' 


*+  Ns  VL'*+  M"2  +  N'~2 ' 


Hence  cos  V  = 


TT.U    i-        u     •       i      r          .\'—mz-\-p]  x  =  mz-\-p'  | 

If  the  lines  be  m  the   form  .        ,   \ 

y  —  nz  +  q    }  y  =  n  z  +  q    \ 

wm+nn'+i 

Then         cos  V  =               ^= —  =..      (41) 

2/2  , 


Ex.   i.   Find  the  cosine  of  the  angle  between  the  lines 

X**  2Z  +  6 


Ex.  2.    Find  the  cosine  of  the  angle  between  the  lines 
x—y'—z     =4)  x+y  +  z   =2p2' 


32  NOTES   ON   SOLID   GEOMETRY. 

These  equations  may  be  put  in  the  forms 


=  —  _=_    (I)  and  ___  = 

4-3  -5 


,T       —  5  +  12  — 
.  •.  cos  V  =  — - 


V26X38 

54.   The  condition  of  perpendicularity  of  two  lines  given  by  the 
equations  in  last  article  is  LL'  +  MM'  +  NN'^  o.     (42) 
The  condition  that  they  shall  be  parallel  (see  Art.  15) 


is  (LM/-L'M)2  +  (LN/-L'N)2+(MN/-M'N)2i=o 

or  -—=—^=—-(43).  These  two  conditions  when  the  lines  are  in  the 

1  A          IV  J.  IN 


x  =  mz+p  )       x  =  m  z+p 
forms 

y  —  nz  +  q   \        y  — 


m'z+p'  \ 
n'z  +  q'   f 


become  mm'  +  nn'+i  •=.  o,  (44)    and  m  ==  m',  n  =  ri    (45)  respec 
tively. 

55.    To  find  the  condition  of  the  intersection  of  two  lines 

x  =  m,+p  } 

V  —  nz  +  q  ) 

This  is  derived  by  eliminating  x,  y  and  z  from  the  four  equations. 
Subtracting  the  third  from  the  first  we  have  o  —(m  —  m'}z  +p—f>'. 

/0  z  —  _P~P  ^    Similarly  from  the  second  and  fourth  z  ~~nr^_n, 
and  since  the  lines  intersect  these  two  values  of  z  are  equal.     There 

fore  we  have  -^^=   ^  .     (46) 
m  —m        n  —n 


Ex.   Find  /  so  that  the  lines  i    ^  ~ 


tersect. 

If  the  two  lines  are  in  the  form 

x—a        y—b       z—c     ,        x— a' __  y— b'  _  z  —  c' 

~TT     TT~    "N~ (I)'  ~T7"=  ~W~'      Nr 


NOTES   ON  SOLID   GEOMETRY.  33 

the  elimination  of  x,  y  and  z  can  be  effected  more  readily  by  writ 
ing  (i)=  K  and  (2)  =  K'. 

.-.  x-a  ^LK    )  , 

r'Vif        .  «  «— <I   =I*K  —  LK. 

x—  a  =  L  K  j 

Similarly  b-  b'  —  M'K'— MK 
f-^r  ^  N'K'-NK. 

Therefore  eliminating  K  and  K'  we  have 

L,    -L',   a-a' 
M,  -M',  £-£' 

(N,    -N',    r-/   =ro 

or 


(47) 

The  Straight  Line  under  Given  Conditions. 

56.   The  equations  of  a  straight  line  parallel  to  one  of  the  coordi 
nate  planes  as  xy  are  z  =  c,   v  =  mx  -\-p. 

The  equations  of  a  straight  line  parallel  to  one  of  the  coordinate 

x  = 

axes  as  z.  are 

y  — 

5  7.    7  'o  find  the  equations  of  a  straight  line,  passing  through  a  given  point. 
If  (x, y,  z')  is  the  point 

x—x'      v—y'      z  —  z'      ,  ft. 
we  have  seen  the  equation  is  — == — =-     '     =—  -^—     (4°) 

or  if  the  equations  are  in 

,     r        x  =  mz+p)  x—x'  —  m(z—z\}  ,     ,      TT          .,.  . 

the  form  *  \    then  }        ,/  \  (49).     Hence  if  the 

y  =  nz  +  q  j  ^r-y   =  n(z—z  )  j  v  v 

equations  of  a  straight  line  contain  only  two  arbitrary  constants,  all 
the  lines  obtained  by  assigning  values  to  these  arbitraries  pass  through 
a  single  point. 

58.  To  find  the  equations  of  a  straight  line  passing  through  two  given 
points  (x ,  y ,  z')  (x",  y",  z")  using  (48)  we  have 

— — ~— — — "—  =  — — — ,  or  dividing  (48)  by  this  to  eliminate 
.L  IVl  JN 

3* 


34  NOTES   ON  SOLID   GEOMETRY. 

L,  M,  and  N  we  have 

x—x'        y—y'        z—z'        . 

:?r=P=-7±7=7w-  <5°> 

If  one  of  the  points  as  (x",y,  z")  be  at  the  origin  then  the  equa 
tions  become 


59.   To  find  the  equation  of  a  straight  line  passing  through  a  given 
point  (x',y,  z'}  and  parallel  to  a  given  straight  line 

x—a        y—b       z  —  c 


L  M 

From  the  first  condition  we  must  have  —-W~  ~         /    =  — — -, — 

-L  M  N 

,  ,  L       M       N 

and  Irom  the  second  condition  -=rT=-rT-.=-:.Tr. 

L       M       N 

Hence  the  required  equation  is 

x — x  __  y—y'  _  z—z1  . 

L  M        ~~N •     w2) 

If  the  equation  passing  through  the  point  x',y',  z  be  of  the  form 

A--JV'  =  m'(z-z')  \  x  ==  mz+p 

i         //         »/•!••  and  the  given  line  be 

v—y   ~  n  (z— z  )    )  y—  nz  +  q. 

Then         «'  =  n  and  »/  =  »/,  and  the  line  will  be 

^2$$Y® 

60.  To  find  the  equations  of  a  straight  line  passing  through  a  given 
point  x',  j'r,  z'  and  perpendicular  to  and  intersecting  a  given  right  line 
x — a  _  y  —  b  _z  —  c 

I  m  n 

The  required  line  by  the  first  condition  will  be  of  the  form 


where  L,  M,  and  N  are  to  be  determined  by  the  conditions 

=  o  (Art.  54) 


NOTES  ON  SOLID  GEOMETRY. 
and 


61.  Ex.  i.  Find  the  equation  of  the  line  joining  the  points  (b,  c,  a) 
and  (a,  c,  It)  and  show  that  it  is  perpendicular  to  the  line  joining  the 
origin  and  the  point  midway  between  these  two  points  ;  and  that  it 

is  also  perpendicular  to  the  lines  x  =y  =  z  and— =—-= -. 

a       b      c 

Ex.  2.  The  straight  lines  which  join  the  middle  points  of  the 
opposite  sides  of  a  tetrahedron  all  pass  through  one  point. 

Take  O  one  of  the  vertices  as  origin  and  OA,  OB,  OC  as  the 
axes  of  x,  j',  s. 

Let  M,  M',  M"  be  the  middle  points  of  BC,  AC  and  OC  respec 
tively,  N,  N',  N"  the  middle  points  of  the  edges  OA,  OB  and  AB 
opposite  to  those  respectively.  Then  to  find  the  equations  of  the 
lines  MN,  M'N',  M"N". 

We  apply  the  equation  — ; -  =  '  ,    '   ,,  •——, — -77-  to   the   points 

x  —x'       y  —y        z  —z 

(M,    N)   (M',  N')  (M",  N")  respectively. 
Let  OA  =  20,  ;  OB  =  2b  ;  OC  =  2c. 
Then  M  is  (o,  b,  c)  and  N  is  (a,  o,  o). 
Hence  the  equation  of  MN  is 

J^~^(l  _.^_ ^      /T\ 

~  —a~~  b  ~  c      ^  ' 

Similarly  the  equation  of  M'N'  is 

x  __y—b  _z         . 
a         —b        c 

And  the  equation  of  M"N"  is 

X          V        Z  —  C 


(i)  and  (2)  give  x  — — ,  y  =—,  z  =  —  and  these  values  satisfy  (3). 
Consequently  these  lines  pass  through  the  point  (-,— ,  - •-). 


36  A'07^S  OA'  SOLID    GEOMETRY. 


Straight  Line,  and  Plane. 

62.  To  find  the  conditions  that  a  line  shall  be  perpendicular  to  a  plane 
given  by  its  equation. 

If  the  plane  be  of  the  form  .v  cos  a+y  cos  fi  +  z  cos  y  =/  (i) 
we  know  that  cos  <v,  cos  /?,  cos  y  are  the  direction  cosines  of  the 
perpendicular  from  the  origin  on  the  plane. 

And  the  equation  of  this  perpendicular  will  be 


cos  a      cos  //      cos  y  ' 

If  any  plane  A^  +  Br  +  Cs=  D  be  parallel  to  the  plane  (r)  we  must 
have 

A  B  C 

cos  a    -  cos//  ~~  cos  y 

and  if  the  line    — — — -~= — —        C-  be  parallel   to  the  line 

x  v  z 

-,  we  must  have 


cos  at       cos  ft      cos  y 

L  M  N 


cos  a      cos/3       cosy 
Hence  the  conditions  that  the  line  —  =  —  —     M    =  -^      shall  be 


perpendicular  to  the  plane  Ax  +  Ey  +  Cz  =  D  will  be 
ABC 

==      (54) 


)  .  x—p      \~q      z 

If  the  line  be  in  the  form  v  we  write  it-        ="—     -  =  - 

)  m          p         i 


A        B        C         A  =  ;;/C) 

And  the  conditions  are  —  -  =  -  —  =  —  or  _,  ,  \    (55) 

m        n         i          B  =  »C  ) 

The  equation  of  a  line  passing  through  the  point  .r',  r',  z1  and 
perpendicular  to  the  plane  Ax  +  fy  +  Cz  =  D  will  then  be 


x—x      y—v       z—z 


A  B 


NOTES  ON  SOLID    GEOMETRY.  37 

or  in  the  unsymmetrical  form 

*-*'  =  ~  (z-z) 

y—y'  =  ~(z-z). 

Ex.  Find  the  equation  of  a  line  passing  through  the  point  (i,  2, 
3)  and  perpendicular  to  the  plane  $x+2j' — \z  =  5. 

63.  To  find  the  condition   that  a   straight  line  shall  be  parallel  to  a 
given  plane.      Let  the   plane  be  Ax  +  l$y  +  Cz  =  D  and   the   line  of 

x — a        y  —  b      z — c 
the  form  —     ^-     ~   . 

Now  if  this  line  is  parallel  to  the  plane  it  will  be  perpendicular  to 
the  normal  to  the  plane.  Hence  the  required  condition  will  be 

AL  +  BM+CN  =  0.      (56) 

64.  To  find  the  conditions  that  a  straight  line  shall  coincide  with  a  given 
plane  Ax  +  By  +  Cz  =  D. 

x  —  a  _  y  —  b  _z  —  c 
L,       TT~  :~N~' 

The  line  must  fulfil  the  condition  (56)  of  parallelism  above, 
AL  -f  BM  -f  CN  =  o.  And  also  any  point  on  the  line  as  (a,  b,  c) 
must  satisfy  the  equation  of  the  plane.  Hence  we  must  have  the 
additional  condition  Aa  +  Bl>  +  Cc— D  =  o.  (57) 

2°.   Let  the  equations  of  the  line  be  of  the  form  x=mz  +p  ) 

y=nz  +  q  }  .  Sub 
stituting  these  values  of  ,r  and  y  in  the  equation  of  the  plane,  we 
have 

A(ms  +/>)  +  K(nz  -f  q)  +  Cz  =  D, 

whence  z  =  —  52 £JT — t     And  for  coincidence  this  value  of 

Am  +  B/*  +  C 

z  must  be  indeterminate,  and  therefore  A/  +  B^  —  D  =  o  )  (58) 

Am  +  B«  +  C  =  o  f  are  the 
conditions  of  coincidence. 

NOTE.     This  last  method  is  a  general  one  of  determining  the  con 
ditions  of  coincidence  of  a  straight  line  and  any  surface  given  by  its 
equation.     That  is  substitute  x  and^  of  the  line  in  the  equation  of 
the  surface  and  since  the  z  in  the  resulting  equation  mu^t  be  inde- 
4 


38  NOTES  ON   SOLID   GEOMETRY, 

terminate  if  there  be  coincidence  we  treat  this  equation  as  an  iden 
tity  and  make  the  coefficients  of  the  different  powers  of  z  separately 
equal  to  zero.  , 

65.   To  find  the  expression  for  the  length  of  the  perpendicular  PD  from 
any  point  P(x',  y',z')  on  a  straight  lin.>  AB  given  by  its  equation. 

i°.  Let  line  be  -      -=  :       -„—  —     —  where  a,  b,  c  are  the  coordi- 
cos  a       cos  p      cos  y 

nates  of  any  point  A  on  the  line.     Now  PD2  =  PA2  —  AD2.     (Fig.  13.) 

But  PA2  =  (x'-ay  +  (v'-by  +  (z'-cY  and  AD  being  the  projec 
tion  of  PA  on  AB,  we  have 

AD  —  (x'  —  a)  cos  a+(y—b)  cos  /S+(z'—6)  cos  y. 
Hence 


+  (z'-b)cosy)\     (59) 
2°.   If  the  given  line  be  of  the  form 

x—  a  __  y—  b  _z—c 

A  B       :    C 

Then 

A 

cos  a  =      ____        —  •  etc.,  etc. 

And  therefore  PD2 


3°.  If  the  given  line  be  x  —  mz  +  p  ) 
y  =  nz  +  q  ) 
Then  PD2 


66.  To  find  the  expressionfor  the  shortest  distance  between  two  straight 
lines  given  by  their  equations, 

This  shortest  distance  is  a  straight  line  AB  perpendicular  to  both 
the  given  lines  PB  and  SR.  (Fig.  14.) 

Let  the  given  equations 

x—  a        v—b       z  —  c         .  x—a       y—b'       z  —  c' 

—  =—  —  =  ---  and  -    —  7—  —  -z  -,=  --  7  and  6  =  the 
co  5  a      cos//     cos;/          cos  a      cos  ft      cos  y 

angle  between  the  lines. 

And  L,  M,  N  the  direction  cosines  of  the  perpendicular  AB. 


NOTES   ON   SOLID    GEOMETRY. 

Then  we  must  have 

L  cos  a  +  M  cos  ft  -f  N  cos  y     —  o  } 
L  cos  a  +  M  cos  ft'  +  N  cos  ;/  =  o.  j 

Whence 

L  M 


39 


cos  /?  cos  T— -cos  /     cos  y       cos  <*  CO3  y  —  cos  a  cos 

N 


cos  <*  cos  ft'  —  cos  M  cos  ft 


M2-|- 


' [(COS /S  COSy'- COS /3' Cos  y)*-t  ^COS  a  COS  >'- CO*  a'  COS  >)  2  -j- (COS  a  COS  ^'-COBa'  COS  0)21 

I 


sn 
(Art.  15). 

Now  let  P  be  the  point  (a,  b,  c)  on  the  line  PB  and  Q  be  the  point 
(a',  b\  c)  on  the  line  SR,  Then  as  the  projection  of  PQ  on  AB  is 
AB  itself,  we  have 

AB  =  (a-a')L  +  (6-6')M  +  (c-c')N  = 

(62) 
(a-  a/)(cos^cosv'_co!»•ycos/3')4-(6—  6')  (cosa'cosy  -  cosacosy  '  )+(c  -c')  (cosacos/3'  -cosa'cos^S) 


If  the  given  lines  are  expressed  in  other  forms  we  can  find  cos  a, 
cos  ft,  etc,  from  the  given  equations  and  substitute  them  in  (62). 


CHAPTER   V. 


TRANSFORMATION   OF  COORDINATES. 

67.  To  transform  to  parallel  axes  through  a  new  origin  the  coordi 
nates  of  which  referred  to  the  old  axes  are  a,  b,  c. 

Let  OA  =  .r,  AN  =  r,  NP  =  z  be  the  coordinates  of  P  referred  to 
the  origin  O  and  the  axes  O,r,  Qv  and  Oz.  Also  let  O'  be  the  new 
origin,  and  OA'  =  a,  A'N'  =  b,  N'O'  =  c  be  its  coordinates  and  let 
O'H  =  x',  HK  =_/  and  KP  =  z  be  the  coordinates  of  P  referred  to 
O'  as  origin  and  axes  parallel  to  the  original  axes.  (Fig.  15.) 

Then  .r  ==  OA  =  OA'  +  A'A 

or  x  =  a  +  x   ) 

Similarly  y  =  b  -f  r'  !•  (63) 

and  z  —  c  +  z'  \ 

Substituting  these  values  in  the  equation  of  a  surface  we  obtain  the 
equation  referred  to  the  new  origin  and  axes. 

68.  To  pass  from  a  rectangular  system  to  another  system  tlie  origin 
remaining  tlie  same. 

Let  Ox,  Or,  Oz  be  the  old  axes  at  right  angles  to  each  other  ;  O_v', 
O/,  Oz'  the  new  axes  inclined  to  each  other  at  any  angle.  (Fig.  16.) 

OM  =  x,  MN  =.y,  NP  =  z 
OM'  =.<  M'N'  =.r',  NT  =  *'. 

Now  the  projection  of  the  broken  line  OM'-f  MN'  +  N'P  on  the 
axis  Ox  is  equal  to  the  projection  OM  of  the  radius  vector  OP  on 
Ox.  Let  cos  a,  cos  #',  cos  a"  be  the  cosines  of  the  angles  which  the 
new  axes  make  with  the  axis  O.*  ;  then 

x  =  x'  cos  a  +  y'  cos  a'  +  z1  cos  a'1 

40 


NOTES   ON    SOLID   GEOMETRY.  41 

If  cos  ft,  cos  ft',  cos  ft"  be  the  cosines  of  angles  which  the  new 
axes  make  with  the  axis  Qy,  and  cos  y,  cos  /,  cos  y",  the  cosines 
of  the  angles  which  they  make  with  Oz,  we  shall  have  similar  values 
and  z.  Hence  the  three  equations  of  transformation  are 

x  =  x'  cos  a  +  y  cos  a'  +  z  cos  a"  j 
j/  =  ^'  cos  ft  +  y  cos  /?'  +  0'  cos  /S"   >     (64) 
z  =  x'  cos  y  +  y  cos  /  +  z'  cos  7".  ) 

We  have  of  course 


cos2  a  +  cos2  ft   +cos2  y    —  i    } 
cos2  a  +  cos2  ft'  +  cos*  y'    =  i    >• 


(B) 


cos 


For  the  angles  A,  /*,  v  between  the  new  axes  of  y  and  z  ',  of  3'  and 
.r',  of  ..v'  and  y  respectively  we  have 

cos  A  =  cos  a   cos  a"  +  cos  /J1'  cos  ft"  +  cos  y'  cos  ;/'    J 
cos  fA  =  cos  «r"cos  or   +  cos  //'cos  /?   +cos  y"cos  y      I  (C) 
cos  r  =  cos  «    cos  a    +  cos  y5  cos  ft'  +cos  ^  cos  y'.    } 

69.  To  pass  from  one  system  of  rectangular  coordinates  to  another  also 
rectangular. 

The  formulae  in  this  case  are  the  same  as  those  in  the  last  with  the 
exception  that  since  the  new  axes  are  also  rectangular  cos  A  =  o,  cos  // 
=  o,  cos  v  =o  and  formulae  (C)  give 

cos  a'  cos  a"  +  cos  ft'  cos  ft"  +  cos  y'  cos  y"=  o   ) 
cos  a"  cos  a  +  cos  ft"cos  ft  +cos  ^"cos  y  =  o    V  (D) 
cos  <*    cos  «'  +  cos  ft  cos  /T  +cos  y  cos  y'  —  o.) 

Since  between  the  nine  quantities  there  are  six  equations  of  con 
ditions,  (B)  and  (D)  there  are  only  three  of  the  quantities,  cos  a, 
cos  ft,  etc.,  independent. 

70.  In  changing  from  rectangular  axes  to  rectangular,   there  is 
another  set  of  equations  of  condition  among  the   quantities,  cos  a, 
cos  ft,  etc.,  equivalent  to  the  preceding  which  result  from  the  fact 
that  the  new  axes  are  rectangular.      For  a,  a',  a"  being  the  angles 
made  by  the  old  axis  of  .v  with  the  new  rectangular  axes,  etc.,  we 
must  have 


cos2  tf  +  cos2  <*'  +  cos2  a"  =  i  } 
cos2  /?  +  cos2  yS'  +  cos2  ft"  =  i  (•    (E) 
cos2  ;/  +  cos2  7/  +  COS2  y"  =  i  ) 


42  A'07v<:s  av  SOLID  GEOMETRY. 


cos  a  cos  /3  +  cos  rr'  cos  ft'  +  cos  <*"  cos  y#"  =  o  i 

cos  a  cos  7  +  cos  <?'  cos  ;/  +  cos  «"  cos   7"  =  o  >-    (F) 

cos  /?  cos  7  +  cos  ft'  cos  7'  +  cos  ft"  cos  7"  —  o  ; 

and  the  new  coordinates  expressed  in  terms  of  the  old  are 

.v'  =  .v  cos  a   +jcos  ft  +scos  y    \ 
y'  —  ^cos  a   +ycosft'  +s  cos  y'    •    (F) 
z   —  x  cos  «"  +y  cos/?"  +s  cos  7"  ) 

71.  In  the  study  of  surfaces  by  sections  made  by  planes  it  is  often 
necessary  to  transform  the  coordinates  in  space  to  coordinates  in 
the  cutting  plane.  To  do  this  we  must  fix  the  plane  with  reference 
to  the  old  coordinate  planes.  Let  the  equation  of  the  plane  be  given 
as  z=  A^  +  By.  Then  the  angle  6  which  this  makes  with  the  plane 

xy  is  determined   bv  the  equation  cos  6  =  —  —  —    and   the 


angle   cp  which   it  traces  on  that  plane  makes    with  the  axis  of  x 

^ 
by  the  equation  tan  cp  =  —  —  ,  the  trace  being  Ajv  +  Bj>  =  o. 

Let  x'Oy'  be  the  given  plane,  cutting  the  plane  xy  in  the  line 
Ox'  which  take  for  the  axis  of  x'  and  let  CV  a  line  perpendicular 
to  it  in  the  given  plane  be  the  axis  of  v'  and  OR  =  .r',  RM  =y 
the  coordinates  of  any  point  M  in  the  plane  referred  to  the  axes  Ox\ 
(V  ;  also  let  OQ  =  .v,  QP  =  r,  P.AI  =a  z  be  the  coordinates  of  M 
referred  to  the  old  axes  O  v,  Or,  Os.  Then  the  angle  MRP  =  9 
and  xOx'  =  (p.  (Fig.  17.) 

Then  PR  =y  cos  ft  PM  =j/  sin  8. 

OQ  =  OR  cos  (p  +  RPsin  <py  QP  =  OR  sin  <p-RP  co^  (p. 

.-.  z-y'  sin  8  \ 

x  =  x'  cos  q>+y'  cos  8  sin  gj  V    (65) 
y  =  .r'  sin  <p—y  cos  8  cos  cp  ) 

And  if  these  values  be  substituted  in  the  equation  of  any  surface 
F(.v,  r,  z)  —  o  the  result  will  be  a  relation  between  x'  andy,  coor 
dinates  of  the  curve  cut  from  the  surface  by  the  plane. 

72.  If  the  cutting  plane  contain  one  of  the  coordinate  axes,  the 
formulae  are  simplified  and  in  many  cases  sufficiently  general. 

Let  x'Oy  (Fig.  18)  be  the  cutting  plane  containing  the  axis  of  y  ; 
Ox'  its  trace  in  the  plane  ;  zx  the  axis  of  x'  ;  PM  =  .v',  OM  =  _r', 


.VOTES   ON   SOLID    GEOMETRY.  43 

the  coordinates  of  any  point  P  in  the  section  ;  ON  =.  x,  NQ  =.r, 
QP  =  2,  the  coordinates  of  P  referred  to  the  old  axes.  Then  angle 
PMQ  =  0,  and  MQ  =  x'  cos  0  PQ  =  x'  sin  0.  .  •,  The  formulae 
of  transformation  are 

x  =  x'  cos  6*   ) 
^=/  •     (66) 

z  —  x  sin  0    } 

That  is,  we  have  only  to  make  x  =  x'  cos  0,  z  —  x  sin  0,  y  -=.y'  in 
the  equation  of  any  surface,  in  order  to  find  the  equation  of  the  sec 
tion  of  this  surface  by  a  plane  containing  the  axis  of j1  and  making  an 
angle  0  with  the  plane  xy. 


CHAPTER    VL 

THE   SPHERE. 

73.   To  find  the  equation  of  the  sphere. 

i°.   In  rectangular  coordinates. 

Let  «,  <£,  c  be  coordinates  of  the  Centre,  and  Radius  =  R. 

The  equation  is  then  (Art.  1  1  )(x—  aY  +  (y-l>Y  +  (z—  c)z—Rz  (67) 
or  if  the  origin  be  at  the  centre 

^+/  +  22  =  R2.      (68) 

2°.  Tn  oblique  coordinates. 

Let  A,  /<,  v  be  the  angles  of  the  axes  then  the  equation  is  (Art.  16) 


x-a)(y-b)  cos  A-f 
2(x—a)(z—c)  cos  )A+2(y—b)(z—c)  cos  r  =  R2     (69) 
or  if  the  origin  be  at  the  centre 

x*  +  .)  2  +  -s2  +  2xy  cos  A  -f  2AT2  cos  yu  +  2yz  cos  r  =  R2.     (70) 

3°.    In  polar  coordinates 

Let  r',  a,  ft  be  the  polar  coordinates  of  the  centre  then  the  equa 
tion  is 


'2  —  2rr'(cos  ^  cos  rt'  +  sin  8  sin  ^  cos  (q>  —  >5))  =  Rtf.      (71) 

If  the  pole  be  at  the  origin  and  the  centre  on  the  axis  of  0,  the 
equation  is 

r=2Rcos#.      (72) 

Since  that  is  the  equation  of  the  generating  circle  in  any  one  of  its 
positions. 

44 


NOTES  ON  SOLID   GEOMETRY. 


74.  The  Sphere  under  conditions  (coordinates  rectangular). 
The  equation  (67)  may  be  written 


45 


JVa  +>*  +  £*  —  2CLX—  2by—  21-2  +  0*  +  &*  +  €*  —  R2    —  O 

or  .T2  -f  /  +  ,s2  +  DA-  +  Ey  +  FZ  +  G  =  o.      (73) 

And  since  this  equation  contains  four  arbitrary  constants,  the 
sphere  may  be  made  to  fulfil  four  conditions  (which  are  compati 
ble)  and  no  more.  Four  given  conditions  give  four  equations  for 
determining  the  constants  D,  E,  F,  G,  and  with  these  determined 
we  know  the  radius  and  centre  of  the  sphere,  for  we  have  only  by 
completing  the  squares  to  throw  the  equation  (73)  into  the  form 

/       DV     /        ^\2     /       F\2       D2     E2     F- 
\x+-\  +  (  y  +  -  )  +(*  +  -)   =  —  +  —  +  --  C* 
\         2/       \         27       V         a/  444 

/     D      E      F\ 
to  see  that  the  centre  is  (  --  ,  --  ,  --  )  and  the  radius  is 

V        2'       2'         2j 

^      F~~ 

7" 

1°.  The  equation  of  a  sphere  passing  through  a  givtn  point  d,  e,  f,  is 
v24-y  +  02  +  D(^-^)  +  E(.y-^)  +  F(0-/)-^-^-/2  =  o.  (74) 
If  the  given  point  be  the  origin  the  equation  is 

=  o.    (75) 


2°.  The  equation  of  a  sphere  cutting  the  axis  of  z  at  distances  c  and  c' 
from  the  origin  is 

x*+^+(z-c)(z-c')  +  I)x  +  Ey=  o     (76)     for"1'  ^ 

must  give  two  values  for  z,  c  and  c  ',  and   this  equation  fulfils  that 
condition. 

3°.  The  equation  of  a  sphere  touching  the  axis  of  z  at  a  distance  cfrom 
the  origin  is 


y  =  o  (77)  for  this  gives  two  coinci 

dent  values  of  z  =  c  when  "          >  . 
J'=o  f 


46  NOTES  ON   SOLID   GEOMETRY. 

4°.    7  he  equation  of  a  sphere  touching  all  three  axes  at  distance  a,  from 
origin. 

To  meet  these  conditions  the  equation  must  be  of  such  a  form  as 

je=o  ) 
to  give  equal  roots  for  z  when  \    the  same  equal  roots  for  y 

when  *          >    and  the   same  equal  roots   for  x  when  -          I  .      Let 

z=o  }  z=o  j 

the  distance  of  points  of  contact  from  origin  be  a,  then  the  equation 
will  be 


=  o     (78) 

as  this  fulfils  the  above  conditions. 

5°.  The  equation  of  a  sphere  passing  through  the  origin  and  having  its 
centre  on  the  axis  of  x  is 

*s+y  +  **=  2R.V.     (79) 

6°.  The  equation  of  a  sphere  tangent  to  the  plane  xy  at  the  point  (a,  b) 
is 

(;t--rt)2-f  0'-£)2  +  *2  +  F*=o  (80) 

for  then  z=o  gives  x=a,  and  y=bt  a  point  (a,  b)  in  the  plane  xy. 
75.    Interpretation  of  the  expression 

(x-a?  +  (.y-b)*  +  (z-c)*-W.     (i) 

i°.  Let  (x,  y,  z)  be  the  coordinates  of  a  point  P  without  the 
sphere  whose  centre  O  is  (a,  b,  c]  and  radius  =  R  and  let  PM  be 
tangent  to  this  sphere  at  the  point  M.  Then  PM2  =  OP2  —  OM2. 


Now 
and  hence  PM2  =  (x-a)*+  (jy-£)2  +  (z-<)2-R2. 

Therefore  the  expression  (i)  is  the  square  of  the  tangent  from  the 
point  P  to  the  sphere. 

2°.  Let  P  (x,  y}  z)  be  a  point  within  the  sphere.  Join  OP  and 
erect  a  perpendicular  PM  to  OP  meeting  the  sphere  in  M,  and  join 
OM. 

Then  PM8=OM*-OP'  =  R*  -((x-ay  +  (y-by  +  (z-c)*} 


NOTES  ON  SOLID   GEOMETRY.  ^ 

That  is  the  expression  (i)  becomes  negative  and  represents  the 
square  of  the  half  chord  through  P  perpendicular  to  the  radius 
through  P. 

76.  .Radical  plane  of  two  spheres. 

Def.    The  radical  plane  of  two  spheres  is  the  plane  the  tangents 
drawn  from  any  point  of  which  to  the  two  spheres  are  equal. 
If  the  equations  of  the  two  spheres  are 


the  equation  of  their  radical  plane  is 


—  o 

For  this  expresses  (Art.  75)  that  the  squares  of  the  tangents  from 
point  (_r,  y,  z)  to  the  two  spheres  are  equal,  and  moreover  it  is  an 
equation  of  the  first  degree  in  x,  y  and  z  and  therefore  the  equation 
of  a  plane.  If  the  spheres  intersect  their  radical  plane  is  their 
plane  of  intersection.  It  may  be  easily  proved  that  the  radical 
plane  of  two  spheres  is  perpendicular  to  the  line  joining  their  cen 
tres. 

77.  ,The    six    radical  planes  of  four  spheres  intersect  in  a  common 
point. 

Let  S  =  o,  S'  =  o  ;  S"  =  o  ;  S'"  —  o  be  the  equations  of  the  four 
spheres.  Then  the  equations  of  their  radical  planes  are 

S-S'  =o  S'  -S"  =o 

S-S"  =0  S'  -  S'"  =  o 

S-S'"  =  o  S"-S"f:==o 

These  may  be  arranged  in  groups  of  four  equations,  which  added 
vanish  simultaneously  and  therefore  the  planes  intersect  in  a  common 
point.  This  point  of  intersection  of  the  six  radical  planes  is  called 
the  radical  centre  of  the  four  spheres. 

78.  Examples  : 

i°.   Find  the  centres  and  radii  respectively  of  the  spheres 


48  NOTES  ON   SOLID   GEOMETRY. 

—  40  =  o. 


—  \x  +  5  jy  =  o. 


2°.  F'ind  the  equation  of  a  sphere  passing  through  the  origin  and 
the  points  i,  2,  3,    —  i,  4,  5,    3,  °>  i- 


CHAPTER     VII. 

CYLINDERS,  CONES,  AND   SURFACES   OF   REVOLUTION. 

79.  CYLINDERS.    Def.  A  cylinder  is  a  surface  generated  by  the  motion 
of  a  straight  line  which  always  intersects  a  given  plane  curve,  and  is 
always  parallel  to  a  fixed  straight  line.     The  moving  straight  line  is 
called  \\\Qgeneratrix;  the  plane  curve  which  it  always  intersects  is 
called  the  directrix  or  guiding  curve. 

80.  To  find  the  general  equation  of  a  cylinder  (coordinates  rectangular}. 
Let  m,  n,   i  be  the  direction  cosines  of  the  axis. 

And  let  >       (i)  be  the  equations  of  the  generatrix  in 

y  —  nz  +  q  \ 

which  m  and  n  are  constant  since  the  generatrix  remains  parallel  to 
the  axis.     For  convenience  take  the  guiding  curve  in  the  plane  xy, 

its  equations  will  then  be  '  ^'~    °  \  .     (2)     Now  making  z  =  o  in 

z — o  5 

the  equations  (i)  we  obtain  x=p  y— q  for  the  point  in  which  the 
generator  pierces  the  guiding  curve  ¥(x,  y)  in  the  plane  xy. 

Hence  we  have  F(/>,  q)  =  o,  (3)  and  eliminating  the  arbitrages/ 
and  q  between  (i)  and  (3)  we  obtain 

?(x-mz,y-nz)  =  o  (82) 

the  general  equation  of  cylinders. 

If  the  cylinder  be  a  right  cylinder  with  its  guiding  curve  in  the 
plane  xy  and  the  axis  of  z  for  its  axis,  then  in  equation  (82)  m  =  o, 
and  »  =  o,  and  the  required  equation  of  the  cylinder  is 

F(*,  y)  =  o.     (83) 


NOTES   ON  SOLID   GEOMETRY. 


49 


8  1.  Cylinders  of  second  order.  We  shall  confine  ourselves  to  cylin 
ders  whose  equations  are  of  the  second  degree. 

i°.   To  find  the  equation  of  the  oblique  cylinder  with  circular  base. 

Here  F(x,y)  =  x2  +  /  —  R2  =  o.  Hence  ¥(x—mz,  y—nz)  =  o 
gives  (x—  mzY  +  (y—  nzf  —  R2  =  o  (84)  the  required  equation. 

2°.  To  find  the  equation  of  the  right  cylinder  with  circular  base.  If 
the  axis  be  the  axis  of  z,  the  equation  is  J?(xy)  =  o  that  is 


3°.   To  find  the  oblique  cylinder  with  elliptical  base.     Let  the  guiding 


curve  in  plane  xy  be    —  +  ^  =  i,      z  =  o. 
a2      b- 

x*       v* 
Then  ¥(Jftjf)  =  —^  +    »—  J  —  °  an^  the  equation  is 


4°.  The  equation  of  the  right  cylinder  with  elliptical  base  whose 

x*     v* 
axis  is  the  axis  of  z  is   F(x,  y)  =  o,   that   is,  —  s  +  ^  =  i. 

5°.  The  equation  of  the  right  parabolic  cylinder  whose  guiding 
curve  is  f  =  ^dx\  z  —  o,  is  f  —  $dx. 

82.  CONES. 

Def.  A  cone  is  a  surface  generated  by  a  straight  line  which  passes 
through  a  fixed  point  and  always  intersects  a  given  plane  curve. 
The  fixed  point  is  called  the  vertex,  the  moving  line  the  generator, 
and  the  given  plane  curve  the  directrix  or  guiding  curve. 

83.  To  find  the  general  equation  of  a  cone. 

Let  the  coordinates  of  the  vertex  be  (a,  b,  c)  the  equation  of  the 

x  —  a        y  —  b      z  —  c 
generator       ^         '~~n~     ~T~    ^   and   take  the   dircctrix   in   the 

plane  (xy)—  its  equation  being  then     ^1'-r)  ~    °  L     (2).     Now  if  we 

z  —  o  ) 

eliminate  w  and  «  by  means  of  the  definition  of  cone  and  the  equa 
tions  (i)  and  (2),  the  resulting  equation  will  be  the  equation  to  the 
cone,  the  locus  of  the  right  line  (i). 

Making  z  =  o  in  (i)  the  values  of*  and  y,  namely,  x  ~  a  ~  mc  1 

y  =  b  —  nc  ) 

which  result  will  be  the  coordinates  of  the  point  in  which  the  gene- 
5 


50  NOTES   ON  SOLID   GEOMETRY. 

rator  meets  the  plane  xy  and  these  will  consequently  satisfy  ¥(x,j') 
=  o  the  equation  of  the  directrix.      We  have  therefore 

F(a  —  me,  b—nc)  =  o    (3).     But  from   (i)    m  =^~   — ,  n  =  - , 

2  —  C  Z  —  C 

and  therefore  (3)  becomes 


z—c  z  —  c 

az  —  ex     bz— cy 


or 

z—c 

the  general  equation  of  cones.     If  vertex  be  on  axis  of  z,  then  a  —  o 

and  b  =  o  and  equation  (85)  becomes  F( —  — ,  —  — )  =  o.    (86) 

\z  —  c     z  —  c  ) 

84.  Cone  with  vertex  at  origin. 

If  the  vertex  of  the  cone  is  at  the  origin  and  the  directrix  in  a 
plane  parallel  to  the  plane  xy,  and  at  a  distance  c  from  it  then  the 

equation  of  the  generatrix  will  be  —  =  —  =  —,  (i)  the  vertex  (o,  o,  o) 

and  the  directrix  will  be      v*»^J  ~       I  .      (2) 

z  =  c  ) 

To  find  the  point  in  which  the  generator  meets  the  directrix  we 
i).   We  thus  get 


Hence  we  have  ¥(mc,  nc)  =  o,  but  m  =  —  ,  and  n  —-  from  (i). 
Therefore 

<?,7r)=°     (B7) 

is  the  equation  required. 

The  equation  (87)  is  a  homogeneous  equation  in  x,y  and  z. 

85.    Cones  of  second  degree.      1°.    The  equation  of  an  oblique  cone  with 
circular  base. 

The  equation  of  the  directrix  is  ~F(x,  y)  =  x2+y—  R'2  =  o. 
Hence 


z—c          z—c  z—c 


NOTES  ON  SOLID   GEOMETRY.  5! 

or  (az  -ex)*  +  (bz—cy}^—  K\z-c}\      (88) 

2°.   To  find  the  equation  of  a  right  cone  with  circular  base,  the  axis  of 
z  being  the  axis  of  the  cone  and  vertex  being  (o,  o,  c).     The  equation 

.    F(*,.>')=-*'+y-Rs=o 
of  the  directrix  is     v 

z—  c. 


Hence 


^  )=o  is  -^  f  7^-R'= 

W  s—  <7  0—  f2       z—c 


R2 
or  jt-2+y=—  (2—  r)2  (89).     This  is  a  cone  of  revolution  about  the 

axis  of  z. 

3°.    The  equation  of  a  right  cone  with  vertex  at  the  origin  and  circular, 
elliptical,  or  hyperbolic  bases. 

The  equations  of  the  circular  base  (directrix)  are 


Hence 

/  ex    cy\  .       fix1       c^y1  2  R2 

\T"'  ~z)~  *~z*         s2""  ""?"       ^9°' 

The  equations  of  the  elliptical  and  hyperbolic  directrices  are 

^  +-£  _  i  =  o  )  and  |2--|  - 1  =  o  \  respective]y. 
z=c]  z=  c) 

Hence  the  cones  are 

x9      f     z2      .     . 
~  i  =oor--+--=-5      (91) 


c*x*      W 


86.  SURFACES  OF  REVOLUTION. 

To  find  the  general  equation  of  a  surface  generated  by  the  revolution  of 
a  plane  curve  generator  about  the  axis  of  z. 

Let  SPi=r  be  an  ordinate  of  the  point  P  to  the  axis  of  z  of  the 


52  NOTES  ON  SOLID   GEOMETRY. 

plane  curve  and  OM  =  x,    MN  =y,  NP  =  z  the  coordinates  of  P. 
Then  SP2  =  ON2  —  OM2  +  MN2,  or  r2  —  .r2+y 

That  is,  the  distance  from  any  point  of  revolving  curve  (gen 
erator)  from  the  axis  of  z  is  r==vC?4j?  (i).  But  r  being  an  ordi- 
nate  of  the  generating  curve  to  the  axis  of  z  we  must  have  by  the 
equation  of  the  curve  in  any  position  r  =  ¥(z)  (2).  Therefore 
eliminating  the  arbitrary  r  between  (i)  and  (2)  we  have 

V^+y  =  F(s)   (93) 

the  required  equation  of  surfaces  of  revolution  about  axis  of  z. 
If  the  curve  revolved  about  the  axis  of  .r  the  equation  is 

Vy +s2  —  F(JC).   (94) 

87.    Surfaces  of  revolution  of  second  order. 

i°.  Equation  of  Cylinder  of  revolution  about  the  axis  of  z.  The 
equation  of  the  revolving  line  is  r  =  a. 


V.      -**+      =  gives  x-+f  =  a\ 

2°.  Equation  of  a  Cone  of  revolution  about  the  axis  of  z,  vertex  at 
(o,  o,  c).  The  equation  of  the  generating  line  is  r  =  m(z—c}. 

Hence  .#2  +  y2  =  m*(z—c)*  (95)  the  required  equation  where  m  is 
the  tangent  of  the  angle  made  by  side  of  cone  with  axis  of  2. 

3°.  Equation  of  the  Sphere.  The  equation  of  the  generating  curve 
a*  or  r  =\/a* — z2. 


Hence 

4°.  Equation  of  the  Surface  generated  by  the  revolution  of  an  ellipse 
about  its  conjugate  axis. 

r*  Z9  02/70  OX 

The  generator  is  -  +  -=  i  or  r~  =  ~(b  —  z  2). 
Hence  the  equation  of  the  surface  is 


+=«-     (96) 
This  is  one  of  the  ellipsoids  of  revolution  called  the  oblate  spheroid. 


NOTES  ON  SOLID   GEOMETRY, 


53 


5  °.    Equation  of  the  Ellipsoid  generated  by  the  revolution  of  an  ellipse 
about  its  transverse  axis  the  {Prolate  spheroid]. 

Take  the  axis  of  x  as  the  axis  of  revolution.     Then  the  equation 


jv      r 
of  the  generator  is  —  +  75  = 


_ 

or  r2  =  —  >(#2—  x*)>      Hence  vj'*  +  s*  —  F(,r)  gives 
a 


v  _  x 

ir**;«  (97) 

the  required  equation. 

88.  Hyperboloids  of  revolution.  Definitions.  When  the  Hyperbola 
revolves  about  its  conjugate  axis  it  generates  the  Hyperboloid  of  revo 
lution  of  one  sheet.  When  it  revolves  about  the  transverse  axis  it  gen 
erates  the  hyperboloid  of  revolution  of  two  sheets. 

i°.   Equation  of  the  Hyperboloid  of  one  sheet.      Let  the  axis  of  z  be 

r2     z2  a2 

the  conjugate  axis  then  -2—73  =  i  or  r2  =  j3(s8  +  £2).     Hence 


2°.   7%f  equation  of  the  Hyperboloid  of  revolution  of  two  sheets.     Take 
the  axis  of  x  as  the  axis  of  revolution.     Then  the  equation  of  the 

a:2      r2  b* 

generator  is  —  —^  =  i    or  r2  =—  (j\r2—  ^2). 

Hence  for  the  equation  of  the  surface  we  have 


A-          ,-        3 

?--?-='•     (99) 

89.  Equation  of  the  Paraboloid  of  revolution  about  the  axis  of\. 

The  equation  of  the  generator  is  r2  —  \dx. 

Hence  the  equation  of  the  Surface  isjy2  +  22  —  \dx.     (100) 

5* 


CHAPTER    VIII. 
ELLIPSOIDS,   HYPERBOLOIDS,  AND  PARABOLOIDS. 

89.   To  find  the  equation  to  the  surface  of  an  Ellipsoid. 

Def.  This  surface  is  generated  by  a  variable  ellipse  which  always 
moves  parallel  to  a  fixed  plane  and  changes  so  that  its  vertices  lie  on 
two  fixed  ellipses  whose  planes  are  perpendicular  to  each  other  and  to 
the  plane  of  the  moving  ellipse,  and  which  have  one  axis  in  common. 

Let  BC,  CA  (Fig.  19)  be  quadrants  of  the  given  fixed  ellipses  traced 
in  the  planes  >;0,  zx  ;  OC  —  c  their  common  semi-axis  along  the  axis 
of  z,  OA  =  a  (on  the  axis  of  x),  and  OB  =  b  (on  the  axis  of  y)  the 
other  semi-axes  ;  QPR  a  quadrant  of  the  variable  generating  ellipse 
in  any  position,  having  i  s  centre  in  OC  and  two  of  its  vertices  in  the 
ellipses  AC,  BC,  so  that  the  ordinates  QN,  RN  are  its  semi-axes ; 
also  let  ON  =  z,  NM  =  x,  MP  =y  be  the  coordinates  of  any  point 
P  in  it  : 

3?         y* 
Then -^  +  .-     =  i.     And  since  Q  is  on   the  ellipse  AC   we 


have     • — .—  =  i -.      Similarly  — --— =  i . 

a  c  u  c 

Hence  eliminating  RN2  and  QN2  we  have 

*       ^L 


-r+-7,-+--r==',     (loi) 
the  equation  to  the  surface. 

90.    To  determine  the  form  of  the  ellipsoid  from  its  equation.     Since  in 

.v2       y2        z* 
the  equation  — y  -f-  — y-  -f  — -  =  i ,  x  can  only  receive  values  between  a 

54 


NOTES  ON   SOLID   GEOMETRY. 


55 


and  —  a,  y  between  b  and  —  b,  and  z  between  c  and  —  c,  the  surface 
is  limited  in  all  directions. 

A2          V2 

If  we  put  z  =  o  we  obtain  —  -f—  r=  i,  for  the  equation   to   the 

trace  on  xy,  which  is  therefore  the  ellipse  AB. 

x*       z1 
If  we  put  y  —  o  we  have  ~--\  —  ^-  =i,  the  ellipse  AC. 


If  we  put  x  =  o  we  have  1—  r  -f  —  =  i,  or  the  ellipse  BC. 

These  three  sections  by  the  coordinate  planes  are  called  the  princi- 
pal  sections,  and  their  semi-axes  a,  6,  c,  are  the  semi-axes  of  the  ellip 
soid  ;  and  their  vertices  the  vertices  of  the  ellipsoid,  of  which  it  has  six. 

If  we  make  z=h  we  have 


the  equations  of  any  section  parallel  to  xv,  which  is  an  ellipse  similar 
to  AB,  since  its  axes  are  in  the  ratio  of  a  to  b,  whatever  be  the  value 
of  h,  and  which  becomes  imaginary  when  h  >  c.  In  the  same  man 
ner  all  sections  parallel  to  xz9  and  yz  are  ellipses  respectively  similar 
to  AC  and  BC.  The  whole  surface  consists  of  eight  portions  pre 

cisely  similar  and  equal  to  that  represented  in  the  figure. 

yA  ^_  j,2      zz 
Cor.   If  £=0  the  ellipsoid  becomes        —  ^  —  t--—  =  i   the   ellip 

soid  of  revolution  about  the  axis  of  z,  Art.  (87),  all  the  sections  of 
which  by  planes  parallel  toyz,  are  circles.  Hence  the  spheroids  may 
be  generated  by  a  variable  circle  moving  as  the  variable  ellipse,  in 
Def.  Art.  (89). 

91.    To  find  the  equation  to  the  hyperboloid  of  one  sheet. 

Definition.  This  surface  is  generated  by  a  variable  ellipse,  which 
moves  parallel  to  a  fixed  plane,  and  changes  so  that  its  vertices  rest 
on  two  fixed  hyperbolas,  whose  planes  are  perpendicular  to  each 
other,  and  to  the  plane  of  the  moving  ellipse,  the  two  hyperbolas 
having  a  common  conjugate  axis  coincident  with  the  intersection  of 
their  planes.  (Fig.  20.) 

Let  AQ  and  BR  be  the  given  hyperbolas  traced  in  the  planes  zx,yz  ; 
OC  =  c  their  common  semi-conjugate  axis  coinciding  with  the  axis 
of  z  ;  OA  =  a,  OB  =  b  the  semi-transverse  axes  ;  QPR  the  generating 
ellipse  in  any  position  having  its  plane  parallel  to  xy,  its  centre  in 


56  NOTES   ON1  SOLID   GEOMETRY. 

OC,  and  its  vertices  in  the  hyperbolas  AQ,  BR,  so  that  the  ordinates 
NQ,  NR,  are  its  semi-axes.  Also,  let  MN  =  x,  MP  =  y,  ON  —  2, 
be  the  coordinates  of  any  point  P  in  the  generating  ellipse  ;  then  the 
ellipse  PQR  gives 

"NQ2  +  NlT2  =  I' 

NQ2      z* 
Also  from  hyperbola  AQ  2  —  "~=i. 

NR2      z" 
And  from  hyperbola  BR       -- ^  —  i. 

Hence, 

f 


A8  V2  Z2 

or  — -  +  ^-—  —  —  -  —  i  (102)  the  equation  to  the  surface. 

92.  To  determine  the  form  of  the  hyperboloid  of  one  sheet  from  its 
equation. 

Since  the  equation  (102)  admits  values  of  x,  y  and  z  positive  and 
negative  however  large,  the  surface  is  extended  indefinitely  on  all 
sides  of  the  origin.  If  we  put  z  =  o  we  obtain 

~—z-  +  ~i  =  i  for  the  trace  on  xy  which  is  the  ellipse  AB.    Similarly 

r2       s2 
the  sections  by  the  planes  xz  and  j'2  are  respectively  ^ -—  i  the 

v2     z' 
hyperbola  AQ,  and1-^  —  —^  =  i   the  hyperbola  BR.      The  ellipse  AB 

and  the  hyperbolas  AQ  and  BR  are  the  principal  sections.  The  sections 
parallel  to  xy  are  all  ellipses  similar  to  and  greater  than  AB.  The 
sections  parallel  to  xz  and  yz  are  hyperbolas  similar  to  the  principal 
sections. 

The  semi-axes  a  and  b  are  called  the  real  semi-axes  of  the  surface 
and  c  the  imaginary  semi-axis,  since  x  =  o  and  y  =  o  give  z  = 
±  c\/ — i.  The  extremities  of  the  real  axes  are  called  the 
vertices  of  the  surface.  The  surface  is  continuous  and  hence  is 
called  the  hyperboloid  of  one  sheet.  The  hollow  space  in  the  inte 
rior  of  the  volume  of  this  hyperboloid  of  which  the  ellipse  AB  is  the 
smallest  section  has  the  shape  of  an  elliptical  dice-box. 


NOTES   ON  SOLID   GEOMETRY. 


57 


Cor.   If  b  =  a   the  equation  becomes = _  =  i   that  of  the 

fl'2         cj 

hyperboloid  of  revolution  of  one  sheet.     Its  sections  parallel   to  xy 
are  all  circles. 

93.    To  find  the  equation  to  the  hyperboloid  of  two  sheets. 

Definition.  This  surface  is  generated  by  a  variable  ellipse  which 
moves  parallel  to  itself,  with  its  axes  on  two  fixed  planes  at  right  an 
gles  to  each  other  and  to  the  plane  of  the  generating  ellipse  and  ver 
tices  in  two  hyperbolas  in  those  planes  having  a  common  transverse 
axis. 

Let  AQ  and  AR  be  the  given  hyperbolas  traced  in  the  planes  zx, 
xy,  OA  =  a  their  common  semi-transverse  axis  along  the  axis  of  x, 
OB  =  b  OC  =  c  the  semi-conjugate  axes  along  the  axes  of y  and  z  ; 
QPR  the  generating  ellipse  in  any  position  having  its  plane  parallel 
loyzt  its  centre  in  Ox,  and  its  vertices  AQ,  AR  so  that  the  ordinates 
QN,  RN  are  its  semi-axes.  Let  ON  =  x,  MN  =j>,  MP  =  z  be  the 
coordinates  of  any  point  P  in  the  ellipse.  (Fig.  21.) 


also  from  hyperbola  AQ        — "—  —  —  i 

c          a 

and  from  hyperbola  AR         — — "—-  =  —  i. 

l>  a1 


Hence 


- •>•-?• -••• 

the  equation  to  the  surface. 

94.  To  determine  the  form  of  the  hyperboloid  of  two  sheets  from  its  equa 
tion. 

The  equation  shows  that  all  values  of  x  between  -f#  and  —  a  give 
imaginary  results,  therefore  no  part  of  the  surface  can  be  situated  be 
tween  two  planes  parallel  toyz  through  A  and  A'  the  vertices  of  the 
common  transverse  axis  ;  but  the  equation  can  be  satisfied  by  values 


58  NOTES  ON  SOLID   GEOMETRY. 

of  x,  >',  z,  indefinitely  great,  therefore  there  is  no  limit  to  the  distance 
to  which  the  surface  may  extend  on  both  sides  of  the  centre. 


V*  Z* 

If  we  make  x-=  o  we  have  ~-  -\ — —  = 

o  c 

for  the  principal  section  by  the  plane  yz.      For  x  =  ±  h  and  h  >  a 

v*         z*         //2 
we  have  '—-  -\ — ^-  —  -  2 i  which  represents  similar  ellipses.     The 

principal  sections  by  the  planes  xy  and  zx  are  AR  and  AQ,  respec 
tively.  For  the  sections  parallel  to  xy  and  putting  z  =  ±  I  we 

have  — ^ =—  =  i  +  —  a  hyperbola  similar  to  AR  with  its  vertices 

in  AQ  and  the  opposite  branch  of  that  hyperbola  and  conjugate  axis 
parallel  to  Oy.  In  the  same  way  the  sections  parallel  to  zx  are  hy 
perbolas  similar  to  AQ  with  vertices  in  AR  and  its  opposite  branch 
and  conjugate 'axes  parallel  to  Qz,  2a  is  the  real  axis  of  the  surface 
and  its  vertices  the  vertices  of  the  surface.  The  axes  2b  and  2c 
are  the  imaginary  axes  of  the  surface  as  it  cuts  neither y  nor  z.  The 
whole  surface  consists  of  two  indefinitely  extended  sheets  perfectly 
similar  and  equal,  separated  by  an  interval.  Hence  its  name. 

Cor.   \ib-c  the  equation  becomes  ~ — — ^ —  =  i   the  equa 
tion  to  the  hyperboloid  of  revolution  about  its  transverse  axis. 

95.  Asymptotic  cones  to  the  two  hyperboloids. 

i°.   The  hyperboloid  of  one  sheet  has  an  interior  asymptotic  cone. 

x*         r2         z* 
Putting  its  equation  — T  +  ^ -j-  =  i   (i)  in  the  form 


4.      -=s  —  /I  ^  --  -   .      (2)       Now  when  z  is    very  great 


,-  is  very  small,  and  hence  the  limiting  form  of  (2)  for  z  increased 


without  limit  is 
v 


—  r  +  ~-  =  —  r    (3)    the  equation  of  an  elliptical  cone    having 

its  vertex  at  the  origin  and  its  elliptical  section  parallel  to  xy. 

Moreover,  this  elliptical  section  is  always  within  the  corresponding 


NOTES   ON  SOLID   GEOMETRY. 


59 


section  of  the  surface  by  the  same  plane.      For  putting  z  —  ±  h  in 
(i)  and  (2)  respectively  we  have 

~-?-  +  '~rr  =  l  +  — r  f°r  tne  section  of  the  surface 
a1          //  c* 

x*         /         /r 

— ^-  +  4^-  =  —7-  for  the  section  of  the  cone. 

a2         b1          c 2 

This  cone  is  asymptotic  to  the  hyperbola. 

2°.    The  hyperboloid  of  two  sheets  has  an  exterior  asymptotic  cone, 

x~          V*          z~ 

Putting  the  equation  — , ^-r r  =  i    (i)  under  the  form 

a2          (j  r 


we  have  as  a  limiting  form  of 


this  equation  when  jy  and  z  increase  without  limit, 

jv2         j2         z~ 

—  —  =  -----  -|  ---  —   (2)  an   elliptical  cone  with   vertex  at  the  origin 

and  with  an  elliptical  section  parallel  to  the  plane  \z.  Moreover, 
this  elliptical  section  is  greater  than  the  corresponding  section  of 
the  surface  by  the  same  plane.  For  putting  x—  ±.h  in  (i)  and  (2) 

i'2          £2          h* 

respectively  we  have     -,-  H  ---  -„-  =  —  5  —  i 
*  2 


* 

•^     "T"55?" 

This  cone  is  asymptotic  to  both  branches  of  the  hyperboloid. 

96.    To  find  the  equation  to  the  elliptic  paraboloid. 

Definition.  This  surface  is  generated  by  the  motion  of  a  parabola 
whose  vertex  lies  on  a  fixed  parabola,  the  planes  of  the  two  parabolas 
being  perpendicular  to  each  other,  their  axes  parallel  and  their  con 
cavities  turned  in  the  same  direction. 

Let  OR  be  a  parabola  in  the  plane  xy,  its  vertex  at  the  origin,  its 
axis  along  the  axis  of  .r,  and  /  its  latus  rectum  ;  RP  the  generating 
parabola  in  any  position  with  its  plane  parallel  to  zx,  vertex  in  OR, 
and  axis  parallel  to  O.v,  and  let  /'denote  its  latus  rectum.  Also  let 
ON  —  x,  NM  =  r,  MP  —  z  be  the  coordinates  of  any  point  P  in  it  : 
also  draw  RM'  parallel  to  Oy.  (Fig.  22.) 


60  NOTES   ON   SOLID   GEOMETRY. 

Then  z*  =  I'.  RM  =  /'.M'N  and  /  =  I'.OM'; 


.'.    '—j-  +  -j-  =  x  (  IO3)  the  equation  to  the  surface. 

97.  To  determine  the  form  of  the  elliptic  paraboloid  from  its  equation. 
Since  only  positive  values  of  x  are  admissible,  no  part  of  the  sur 

face  is  situated  to  the  left  of  the  planers.  But  the  surface  extends 
indefinitely  in  the  positive  direction  of  x.  If  we  makej'  =  o,  22=  I'x 
is  the  equation  to  the  principal  section  OQ,  and  all  sections  parallel 
to  zx  are  parabolas  equal  to  OQ,  with  vertices  in  OR  ;  similarly,  all 
sections  parallel  to  xy  are  parabolas  equal  to  the  other  principal  sec 
tion  OR,  with  vertices  in  OQ.  If  we  make  x  —  h  we  have 

21  +  Jl-, 
Ih  +  I'h  ~ 

Therefore  the  sections  parallel  to  zyare  similar  ellipses,  and  hence  its 
name. 

Cor.  If  /=  /  the  equation  becomes  j'2+£2  =  /.v,  the  paraboloid  of 
revolution. 

98.  To  find  /he  equation  to  the  hyperbolic  paraboloid. 

Definition.  This  surface  is  generated  by  the  motion  of  a  parabola 
whose  vertex  lies  on  a  fixed  parabola,  the  planes  of  the  two  parabolas 
being  perpendicular  to  each  other,  their  axes  parallel,  and  their  con 
cavities  turned  in  opposite  directions.  (Fig.  23.) 

Let  OR  be  a  parabola  in  the  plane  of  xy,  vertex  at  the  origin,  and 
axis  along  with  the  axis  of  x,  and  /  its  latus  rectum,  RP  the  generat 
ing  parabola  in  any  position,  vertex  in  OR,  axis  parallel  to  O.v, 
and  let  t  denote  its  latus  rectum,  and  ON  =  x,  NM  =y,  MP  —  z, 
the  coordinates  of  any  point  P  in  it  ;  draw  RM'  parallel  to  Oy. 
Then 

z-=!'.  MR  and/=/.OM'; 

but  OM'-  MR  =  ON  =  AT. 

V2  22 

Hence  :    —  ---  -  =  x  (104),  the  equation  of  the  surface. 

99.  To  determine  the  form  of  the  hyperbolic  paraboloid  from  its  equation. 
The  surface  cuts  the  coordinate  axes  only  at  the  origin,  and  since 

the  equation  admits  positive  and  negative  values  of.r,  v,  s,  as  great 


NOTES   ON   SOLID   GEOMETRY.  6f 

as  we  please,  the  surface  extends  indefinitely  both  ways  from  the 
origin. 

If  we  makejy  =  o  we  have  z~  '  =  I'x  the  principal  section,  the  para 
bola  OQ,  with  its  concavity  turned  towards  the  left  ofyz,  and  all  sec 
tions  parallel  to  zx  are  parabolas  equal  to  OQ  with  their  vertices  in 
OR.  Making  z  =  o  we  have  y2  =  Ix  the  parabola  OR,  and  sections 
parallel  to  xy  are  parabolas  equal  to  OR  with  vertices  in  OQ. 

If    we    make    x  —  o    we    have    the     principal    section    in   yz, 

z^  I  =  ±y  +Jl'  or  two  straight  lines  through  the  origin;  and  for  sec 
tions  parallel  toyz  making  x  =  h  we  have 


'-—  --  —f—  =  i   a   hyperbola   with  its  vertices   in  OR,  and  con 
jugate  axis  parallel  to  Oz.      For  h  negative  the  section  becomes 

£2  y2 

—7-  —  '——-  —i  a  hyperbola  with  its  vertices  in  OQ,  and  conjugate 
In        Ik 

axis  parallel  to  O.v. 

The  surface  has  but  one  vertex,  and  consists  of  one  sheet  and  one 
infinite  axis. 

100.   Asymptotic  planes  to  the  hyperbolic  paraboloid. 

v*       z* 
The  equation   —  ---  -jr  =  x  may  be  written 

j'2       z*    /       l'x\ 
z-j~==~jr  (  T  +  ~7  )  which   has  for  its  limiting  form 

/        /     \        z  J 

y'l  Z<1 

when  y  and   z  become  infinitely  great  with  regard  to  x,  -—  =  —7-  , 

V  z  v  z 

or  -:—=  =  ±-——.     This  represents  two   planes    —  =.  =  H  --  —  and 


V  ^ 

-'—=  =  --  -  .  through  the  origin  and  asymptotic   to   the    surface. 

V? 


These  planes  contain  the  asymptotes  to  all  the  hyperbolic  sections  of 
the  surface  parallel  toj'2. 

101.  The  elliptic  and  hyperbolic  paraboloids  are  particular  cases  of  the 
ellipsoid  and  hyperboloid  of  one  sheet  respectively  when  the  centres  of  these 
surfaces  are  removed  to  infinite  distance. 


Take  the  equation  —  +  ^—-  H  —  —  =i,  and  transfer  the  origin  to 


62  NOTES  ON  SOLID    GEOMETKY. 

the   left  vertex  of  the  axis  20,  (—a,  o,  o).      (New  coordinates  being 
parallel  to  the  primitive.) 


-,.  or 

-  1  .     v  J  1  o 

a2 


r2         y* 


or  multiplying  through  by  a    -  --  t-  -y^-  ±  —  =2jr  (i), 


in  which  — ,  and    —  are  the  semi-latera  recta  of  the  principal  sec- 
a  a 

V1  c* 

tions    in  xy   and    zx.     Now  make  a  •=.  oo    and  put   - —  and    — , 

a  a 

which  remain  finite,  equal  to  /  and  /'  respectively. 
.*.  (i)  becomes 

=y-  ±  —JT  —  2x,  the  equations  to  the  paraboloids. 

1 02.  The  equations  of  the  surfaces  of  the  second  order  which  we  have 
been  studying  are  of  the  two  forms 

•=D    (i) 

»=kA*    (2) 

and  we  will  show  hereafter  that  all  the  surfaces  of  the  second  degree 
may  by  transformation  of  coordinates  be  included  in  these  two  forms. 

The  first  form  (i)  includes  the  sphere,  ellipsoid,  hyperboloids, 
cones  of  second  order,  elliptical  and  hyperbolic  cylinders — which 
have  centres.  For  if  — .r,  — -y,  —z  be  written  for  (x,  y,  z)  in  (i) 
the  equation  is  not  altered,  therefore  for  every  point  P  (x,  y,  z)  on 
the  surface  there  is  a  point  P'  (—  x,  —y,  —z)  and  PP'  passes  through 
the  origin  O  and  is  bisected  in  O. 

Moreover,  the  coordinate  planes  bisect  all  the  chords  parallel  to 
the  axes  perpendicular  to  these  planes  respectively  and  are  principal 
planes  of  the  surface. 

The  second  form  (2)  includes  the  elliptic  and  hyperbolic  parabo 
loids  and  the  parabolic  cylinder  which  have  a  centre  at  an  infinite 
distance. 

The  planes ^.s:  and  zx  are  principal  planes  of  the  two  paraboloids, 
the  other  principal  plane  being  at  an  infinite  distance. 

Also  both  families  may  be  represented  by  the  equation 


NOTES  ON   SOLID   GEOME'IRY.  63 


the  origin  being  at  the  vertex  and  A  =  o  when  the  surfaces  have 
no  centre. 

EXAMPLES. 

1.  Construct  the  sphere  whose  polar  equation  is 

r  =  a  sin  6  cos  cp. 

2.  Find  the  locus  of  the  point  the  sum  of  the  squares  of  the  dis 
tances  of  which  from  n  fixed  points  is  constant. 

3.  Find  the  locus  of  the  point  the  ratio  of  the  distances  of  which 
from  two  fixed  points  is  constant. 

4.  Find  the  equation  of  the  surface  generated  by  the  motion  of  a 
variable  circle  whose  diameter  is  one  of  a  system  of  parallel  chords 
of  a  given  circle  to  which  the  plane  of  the  variable  circle  is  perpen 
dicular. 

5.  The  sphere  can  be  represented  by  the  simultaneous  equations 

x  =  a  cos  cp  cos  6  } 
y  =  a  cos  cp  sin  6  >  • 
z  =  a  sin  cp 

6.  The  ellipsoid  may  be  represented  by  the  equations 

x  =  a  cos  <p  cos  6  } 
y  —  b  cos  cp  sin  6  Y  • 
z  =  c  sin  cp 

7.  The  hyperboloid  of  one  sheet  may  be  represented  by  the  equa 
tions 

x  —  a  sec  cp  cos  6  \ 
y  —  b  sec  cp  sin  6  \  - 
z  =  c  tan  gj 

8.  The  hyperboloid  of  two  sheets  may  be  represented  by  the  equa 
tions 

x  —  a  sec  cp  \ 

y  =  b  sin  6  tan  cp  >  • 
z  —  c  cos  6  tan  cp  ) 

9.  A  line  moves  so  that  three  fixed  points  on  it  move  on  three 
fixed  planes  mutually,  at  right  angles.     Find  the  locus  of  any  other 
point  P  on  its  line. 


64  NOTES  ON   SOLID   GEOMETRY. 

Solution  : 

Let  the  three  fixed  planes  be  the  coordinate  planes  (x,y,z)  the  coordinates 
of  P.  A,  B,  C  the  points  in  which  the  line  meets  the  coordinate  planes  of  yz, 
xz,  xy,  respectively.  Take  PA=<z,  PB=/>,  PC=c,  ON=.r,  NQ=,y,  QP  =  z, 
<ACA'=:<p,  <CB'.r=:0  (CA'  being  the  projection  of  CA  on  the  plane  xy  and  B' 
the  projection  of  B  on  the  axis  of  x). 

Then  x=a  cos  cp  cos  Q,y=b  cos  q>  sin  6,  z= c  sin  q>, — and  therefore  the  sur 
face  is  an  ellipsoid. 

10.  Find  the  locus  of  a  point  distance  of  which  from  the  plane  xy 
is  equal  to  its  distance  from  the  axis  of  z  (coordinates  rectangular). 

11.  Find  the  locus  of  the  centres  of  plane  sections  of  a  sphere 
which  all  pass  through  a  point  on  the  surface. 

12.  Find  the  equation   of  the  elliptical   paraboloid  as  a  surface 
generated  by  the  motion  of  a  variable  ellipse  the  extremities  of  whose 
axes   lie  on   two  parabolas  having  a  common  vertex  and  common 
axis  and  whose  planes  are  at  right  angles  to  each  other. 

13.  Find  the  equation  of  the  hyperbolic  paraboloid  as  generated 
in  a  similar  manner  by  the  motion  of  a  variable  hyperbola. 

14.  Construct  the  surface  r  sin  6  =  a. 

15.  Find  the  equation  to  the  surface  B  =  JTT  in  rectangular  coor 
dinates. 


CHAPTER  IX. 
RIGHT  LINE  GENERATORS  AND  CIRCULAR  SECTIONS. 

103.  SURFACES  of  the  second  degree  admit  of  another  division,  viz. 
into    those  which   can    be  generated  by  the  motion    of  a    straight   line 
and  into  those  which  cannot.      This  property  which  we  have  seen   to 
belong  to  the  cylinder  and  cone  we  shall  now  show  to  belong  also  to 
the  hyperboloid  of  one  sheet  and  the  hyperbolic  paraboloid.     The 
ellipsoid  being  a  closed  finite  surface  does  not  possess  this  property ; 
nor  the  hyperboloid  of  two  sheets,  since  that  consists  of  two  surfaces 
separated  by  an  interval ;  nor  the  elliptical  paraboloid,  since  that  is 
limited  in  one  direction. 

104.  Straight  line  generators  of  the  hyperboloid  of  one  sheet. 
The  equation  of  the  hyperboloid  of  one  sheet 

a2          y2          z*  .r2          z*  r 

— v-  +  ^ 5-  —  i      mav  be  written          - -r-  =  i  — ~-~ 

az          o*          c  a2          c  '   b1 

fx    z\  fx    z\  _  f     y\(     y\  -      ( \\ 

\a        c  )  \a        c  )        \  b )  \  b) 

Now  (A)  is  satisfied  by  the  pair  of  equations 

(B) 

Z  f  V\ 

+  7  =KI  +  i)J 

and  also  by  the  pair 


66  A'OTES   ON   SOLID    GEOMETRY. 

And  m  being  arbitrary  equations  (B)  represent  a  system  of  straight 
lines,  and  all  of  these  lie  on  the  hyperboloid  as  the  two  equations 
together  satisfy  the  equation  to  the  hyperboloid. 

Similarly  equations  (C)  represent  another  and  distinct  system  of 
straight  lines  which  also  lie  on  the  hyperboloid  which  is  the  locus  of 
both  systems,  and  we  shall  see  the  lines  of  either  system  may  be  used 
as  generators  of  the  surface. 

105.    No  tivo  generators  of  the  same  system  intersect  one  another, 
For  example  take  two  of  the  system  (B), 

/  /-v        A  v  1 

m'i )  =  i  -  -. 

\a         c  J  b 

(0 
x 

a 


(2) 
j 

a      c  \  j 

Combining  the  first  equation  of  (i)  with  the  first  of  (2)  we  obtain 


(in   —  m")  (  I  —  —  J  = 


Combining  the  second   equation  of  (i)  with  the  second  of  (2)  we 
have 


or  v  ~  —    . 


These  values  for  y  being  incompatible  the  lines  do  not  intersect. 

1 06.   Any  generator  of  the  system  (B)  will  intersect  any  generator  of 
the  system  (C). 

Take 

,„'(*--*-}=  t-Z 
c  /  b 

of  system  (B) 


(x        z  \  ,  (         y\ 

t+r)  =  w('+i)  J 


NOTES   ON   SOLID   GEOMETRY.  67 

U  (4)  of  system  (C). 


Eliminating  x,  y,  and  a;  we  obtain  the  identity  m'm"  —  m'm", 
therefore  the  lines  intersect. 

Hence,  through  any  point  of  an  hyperboloid  of  one  sheet  two 
straight  lines  can  be  drawn  lying  wholly  on  the  surface. 

107.  No  straight  line  lies  on  an  hyperboloid  which  does  not  belong  to 
one  of  the  systems  of  generating  lines  (B)  or  (C). 

For,  if  possible,  suppose  a  straight  line  H  to  lie  entirely  on  the 
hyperboloid,  it  must  meet  an  infinite  number  of  generating  lines  of 
both  systems  (B)  and  (C).  Let  two  of  these  (one  of  B  and  one  of 
C)  intersect  H  in  two  different  points,  \ve  could  then  have  a  plane  in 
tersecting  the  surface  in  three  straight  lines,  which  is  impossible  since 
the  equation  is  of  the  second  degree.  Hence  no  such  line  as  H  can 
lie  on  the  surface. 

1 08.  The  hyperboloid  of  one  sheet  may  be  generated  by  the  motion  of  a 
straight  line  resting  on  three  fixed  straight  lines  which  do  not  intersect,  and 
which  are  not  parallel  to  the  same  plane. 

In  the  first  place  it  is  necessary  that  the  motion  of  a  right  line 
which  is  to  generate  a  surface  should  be  regulated  by  three  condi 
tions.  For,  since  its  equations  contain  four  constants,  four  condi 
tions  would  fix  its  position  absolutely  ;  with  one  condition  less  the 
position  of  the  line  is  so  far  limited  that  it  will  always  be  on  a  certain 
locus  whose  equation  can  be  found. 

Take  then  three  fixed  generating  lines  of  the  system  (B),  these  do 
not  intersect,  nor  are  they  parallel  to  the  same  plane.  Now,  if  a 
straight  line  move  in  such  a  manner  as  always  to  intersect  these  three 
straight  lines,  it  will  trace  out  the  hyperboloid  of  which  they  are  the 
generating  lines. 

For  the  moving  line  meets  the  hyperboloid  in  three  points  (one 
on  each  of  the  fixed  straight  lines),  and  hence  must  necessarily  lie 
wholly  upon  the  surface.  For  the  equation  of  intersection  of  a  line 
and  this  surface  being  a  quadratic  equation,  if  satisfied  by  more  than 
two  roots,  it  is  satisfied  by  an  infinite  number.  The  moving  straight 
line,  therefore,  in  its  different  positions,  will  generate  the  hyper 
boloid. 


68  NOTES   ON  SOLID   GEOMETRY. 

109.   Lines  through  the  origin  parallel  respectively  to  generators  of  the 
systems  (B)  and  (C)  lie  on  the  cone 

x*     y      z- 

— r  +  -73-  =  =  —    asymptotic  to  the  hyperboloid. 
For  this  equation  of  the  cone  may  be  put  in  the  form 

A    A  A  +  A  =  _,vz, 

\  a        c  J     \a        c  t  b      b 

which  gives  two  systems  of  lines  through  the  origin  lying  on  the 
cone,  one  system  evidently  parallel  to  the  lines  (B)  and  the  other  to 
the  lines  (C). 

no.    The  projection    of  a  generating  line  of  either  system    upon  the 
principal  planes.,  is  tangent  to  the  traces  of  the  surface  on  those  planes. 

The  equation  of  the  trace  of  the  surface  on  the  plane  zx  is 


The  projection  of  the  line  of  system  (B)  on  xz 

„»  f£_  i\  +  *  +  £  =2m.  or  ^±}.  *  +  1=*.  i  =  I  (I). 

\    a        c  J        a        c  2m        a          2m      c 

"V  P 

Now.  the  condition  that  a  line  in  the  form   —  ~| —  =  i    shall   be 

p    q 

.r2         z1                   a*        c* 
tangent  to  the  hyperbola    ~ F=I    'IS    ~^~ T— T- 

This -condition  is  fulfilled  by  the  projection  (i),  for 


(«'+!)'      (i-m'Y 

Hence  this  projection  is  tangent  to  the  hyperbola. 

III.    The  straight  line  generators  of  the  hyperbolic  paraboloid. 
The  equation  of  the  hyperbolic  paraboloid 


NOTES   ON  SOLID    GEOMETRY.  69 


•=y  ---  —  =  x  may  be  written 


And  hence  it  is  satisfied  by  the  pair  of  equations 

y 
V7 

» 


n+^)=i  _ 

or  by  the  pair 

y  z  •} 


HE). 


Hence  the  surface  has  two  systems  of  straight  line  generators  (D) 
and  (E). 

The  lines  of  both  systems  are  parallel  to  the  asymptotic  planes 
of  the  surface  respectively.  The  equations  of  these  planes  being 


z 
=•  =  o  and 


112.  We  can   show  in  the  same  manner  as  in   the  Articles  (34) 
and  (35)  that  no  two  lines  of  the  same  system  intersect  ;  and  that  a 
line  of  either  system  intersects  all  the  lines  of  the  other  system,  and 
that  no  other  line  than  the  lines  of  these  two  systems  can  lie  on  the 
hyperbolic   paraboloid.     And  hence  that  through  every  point  of  the 
surface  two  lines  may  be   drawn  which   lie  wholly  on   the  surface. 
And  as  in  (108)  that  this  paraboloid  may  be  generated  by  the  motion 
of  a  straight  line  which  rests  on  two  fixed  straight  lines  and  is  con 
stantly  parallel   to  a  fixed  plane ;  also  by  a  straight  line  which  rests 
on  three  fixed  straight  lines  which  are  all  parallel  to  the  same  plane. 

113.  The  projections  of 'the  generating  lines  on  the  principal  planes  are 
tangent  to  the  principal  sections  of  the  paraboloid. 

The  principal  section  in  xy  is  y*  =  Ix  (i). 


NOJ^ES   ON  SOLID  GEOMETRY 
The  projection  of  any  line  of  the  system  (D)  on  xy  is 


i 

m  2  2m 


—j=  —  mx  ^ or  y  =  -      —  jp-fJ— .     (2) 

/t  /     /  -WJ  O  ^-rw  *       / 


Now  the  tangent  line  to  the  parabola  y>=  Ix  is  of  the  form 

y  —  tx  +  • — :  and  if  /  =  — — —  then  —  =  -5— -. 
4^  2  4/          2m 

Hence  the  projection  (2)  is  tangent  to  the  section  y  =  lxt 

114.  Distinctions  of  surfaces  of  second  order  generated  by  straight 
lines. 

All  the  generators  of  the  cone  intersect  in  one  point.  All 
the  generators  of  the  cylinder  are  parallel.  Hence  cones  and 
cylinders  are  called  developable  ruled  surfaces.  In  the  case  of  the 
hyperboloid  of  one  sheet  and  the  hyperbolic  paraboloid,  the  gen 
erators  of  neither  system  intersect  or  are  parallel.  These  are 
styled  skew  ruled  surfaces.  The  distinction  between  these  last  two 
surfaces  is  that  the  generators  in  the  paraboloid  are  parallel  to  a  fixed 
plane. 

115.  Plane  sections  of  surf  aces  of  the  second  order. 

If  we   intersect  the  surfaces  represented  by  the  general  equation 

A*2  +  By2  +  Os2  +  2h.'xz  +  2B>  +  2Cxy  +  2&'x+2'B"y+2C"z  ••-  D 
by  the  plane  z  =  o  we  will  obtain 

A^2  +  B|'2  +  2C'A7'|-2A"jt:  +  2B''y  =  D  (i)  a  conic  section. 

If  we  intersect  it  by  a  plane  z  —  a  we  have  for  the  curve  of  inter 
section 

Aa2  -h  B^2  +  2C'xy+  2G'x  +  2H>=D', 

a  conic  similar  to  the  conic  (i). 

Therefore  sections  of  surfaces  of  the  second  order  by  parallel 
planes  are  similar  curves,  and  hence,  in  determining  the  form  of  these 
sections  we  may  confine  ourselves  to  the  discussion  of  sections  through 
the  origin. 


NOTES  ON   SOLID   GEOMETRY.  ^ 

1  1  6.   To  determine  the  nature  of  the  curve  formed  by  the  intersection  of 
a  surface  of  the  second  order  by  any  plane. 

Take  the  equation 


=  2Ar,r.     And  in  order  to  get  the  equation  of  the 
curve  of  intersection  in  its  own  plane 

Make 

x  —  x  cos  q>  -f  y  cos  6  sin  cp 

y  zz:  x'  sin  cp  —y  cos  6  cos  cp 
z  =yf  sin  6.  See  Art.  (71). 

Arranging  the  result  we  have 

x'*(K  cos2r/>  +  B  sin2  cp)  +  2x'y'(K  —  B)  cos  6  sin  cp  cos  cp 

+/2((A  sin2  v  +  B  cos2  93)  cos'2  0  +C  sin2  0)  =  2AV  cos  <p 

-f  2Aj/  cos  6  sin  (£>, 

the  equation  to  a  conic  section  which  will  be  an  ellipse,  parabola  or 
hyperbola,  (including  particular  cases  of  these  curves,)  according 
as  the  quantity 

(A—  B)2  cos2  6  cos2  cp  sin2  cp—  (A  cos2  <p  +  B  sin2  <p)(A  cos2  6  sin2  cp 

+  B  cos2  #cos2  <p  +  Csin2  0) 


or        -AB  cos2  cp-AC  cos2  ^  sin2  0—  BC  sin2  <p  sin2  8,      (i) 
is  negative,  zero  or  positive. 

Hence  every  section  of  an  ellipsoid  is  an  ellipse  because  A,  B  and 
C  are  all  positive. 

The  sections  of  the  hyperboloids  may  be  ellipses,  parabolas  or 
hyperbolas  since  one  or  two  of  the  quantities  A,  B  and  C  will  then 
be  negative. 

For  paraboloids  A  =  o.  Hence  for  the  elliptic  paraboloid  in 
which  B  and  C  have  the  same  signs  the  section  is  an  ellipse  ;  except 
when  B  =  o  or  cp  =  o  in  which  cases  it  is  a  parabola. 

For  the  hyperbolic  paraboloid  since  B  and  C  are  of  contrary  signs 
the  section  is  a  hyperbola  except  when  6=  o  or  cp=o  when  it  is  a 
parabola. 

1  1  7.  Circular  sections.  Since  the  section  is  referred  to  rectangular 
axes  it  cannot  be  a  circle  unless  the  coefficient  of  xy'  vanishes 


72  NOTES   ON   SOLID   GEOMETRY. 

or  (A  — B)  cos  6  sin  cp  cos  cp  —  o 

7T  7T 

or  #  —  —  or  &=  - — ,  or  cp  —  o 

2  2 

which  shows  thaty^r  #  circular  section  the  cutting  plane  must  be  perpen 
dicular  to  one  of  the  principal  planes  of  the  surface, 

1 1 8.  Let  us  now  examine  the  surfaces  of  the  second  order  for  cir 
cular  sections. 

Take  first  the  surfaces  having  a  centre  and  therefore  represented 
by  the  equation 

A*2  +  B/  +  Cz2=  i.     (i) 

Since  every  circular  section  must  be  perpendicular  to  a  princi 
pal  plane,  let  the  cutting  plane  contain  the  axis  of  y,  and  make 
the  angle  6  with  the  plane  xy— 

To  transform  (i)  to  this  plane  make 
x  =  x'  cos  6 

y=y 

z  —  x  sin  6.  Art.  (72). 

Hence  we  have 

.v"(A  cos2  0  +  C  sin2  0)  +  B/*  =  i     (2) 
which  represents  a  circle  if 

Acos2#-fC  sin2  0—  B 
B-A 


or 


tan2  9  =  .     (3) 


We  must  now  examine  for  each  of  the  surfaces  which  axis  it  is 
that  coincides  with  the  axis  ofj>. 

i°.  For  the  ellipsoid  A  =  — ,  3=-^-,  C  =— 


Hence  for  a  real  B  b  must  lie  (in  value)  between  a  and  c  or  the 
axis  of  the  surface  to  which  the  cutting  plane  of  circular  sections  is 
parallel  is  its  mean  axis. 

2°.   For  the  hyperboloid  of  one  sheet  since  we  cannot  have  B  ne- 


NOTES   ON   SOLID    GEOMETRY. 


gative  we  must  put  A  =  —^  B  =  —  C  —  -   — 


.'.   b  >  a  or  the   cutting  plane  is  parallel  to  the  greater  of  the  real 
axes. 

3°.   For  the  hyperboloid  of  two  sheets  since  we  cannot  have  A  and 
C  negative,  we  must  put 


.*.   b  >  c  or  the   cutting   plane  is  parallel  to  the  greater  of  the  im 
aginary  axes. 

Since  tan  6  has  two  equal  values  the  cutting  plane  may  be  inclined 
at  an  angle  6  or  180°—  6  to  the  plane  of  xy.  Hence  there  are  two 
sets  of  parallel  circular  sections  of  the  surfaces  having  a.  centre.  If 
the  surface  becomes  one  of  revolution  we  have  tan  6  =  oo  or  o,  and 
the  two  positions  of  the  circular  sections  coincide  with  each  other, 
and  are  parallel  to  the  two  equal  axes. 

119.  Secondly.  For  the  surfaces  not  having  a  centre,  we  take 
equation  B_/  +  Cs2  =  2k'x  (i). 

i°.  For  the  elliptic  paraboloid,  B  and  C  have  the  same  sign. 
Transforming  (i)  we  have  B/2  +  Or'2  sin2#  —  2K'x  cos  6\  and  hence 
for  circular  sections  we  must  have  the  condition  C  sin20  =  B,  or 


=  ±  \  - 


sin  6  =  ±  \  -    Therefore  the  cutting  plane  is  perpendicular  to  the 
^ 

principal  section  whose  latus  rectum  is  least. 

2°.  For  the  hyperbolic  paraboloid,  since  B  and  C  have  different 
signs,  sin  6  is  imaginary,  and  no  plane  can  be  drawn  which  shall  in 
tersect  it  in  a  circle.  This  was  evident,  too,  from  the  fact  (Art. 
116)  that  the  hyperbolic  paraboloid  can  have  no  elliptic  sec 
tions. 

i  20.  Then,  to  sum  up,  all  the  surfaces  discussed  with  the  excep 
tion  of  the  hyperbolic  paraboloid  admit  of  two  sets  of  planes  of  cir- 

7 


74 


NOTES   ON   SOLID    GEOMETRY. 


cular  sections.     Therefore  they  can  be  generated  by  the  motion  of  a 
variable  circle  whose  centre  is  on  a  diameter  of  the  surface. 

121.   The  planes   of  circular  section  may  be  found  directly  from 
the  equations  of  the  surfaces,  as  follows: 

The  equations  of  the  central  surfaces 


may  be  written 
or 


which  shows  that  either  of  the  planes 


<\/A—  BJI'+V^-^—  =  o  (i)  <v/A  —  B.r—  VB—  C.s  =  o  (2) 
cuts  the  surface  in  the  same  line  in  which  it  cuts  the  sphere 


Hence  the  planes  (i)  and  (2)  and  all  planes  parallel  to  them  cut 
the  surface  in  circles. 

The  equation  to  the  elliptic  paraboloid  may  be  treated  in  a  similar 
manner,  thus  showing  its  planes  of  circular  section. 

122.  Sections  of  Cones  and  Cylinders. 

i°.  The  sections  of  the  cones  may  be  inferred  from  Art.  95.  For 
elliptic  cones,  sections  of  the  hyperboloids  by  any  plane  are  always 
similar  to  the  sections  of  the  asymptotic  cone  to  the  surface  made  by 
the  same  plane,  as  is  evident  from  the  equations  respectively.  Hence 
the  section  of  a  cone  of  revolution  by  a  plane  will  give  an  ellipse, 
parabola,  or  hyperbola.  But  we  will  examine  this  case  more  par 
ticularly. 

In  the  equation  of  the  cone  of  revolution 

A.  +  /=  L2  (2  _,  )«,  or  .v>+y  =  (*-<•)' 


(when  -  —          -  J  put  x=x'  cos  0} 

tan  v  J 


NOTES   ON   SOLID    GEOMETRY,  75 

And  we  have  for  the  curve  of  intersection  by  the  plane   containing 
the  axis  ofjy 

jr'2(cos2  #tan2  v  —  sin2  6)  +  _/*  tan2  v+2cx'  sin  6—c1  =  o  (i). 

This  equation  (i)  represents  an  ellipse,  parabola,  or  hyperbola, 
according  as  cos2  6  tan2  v  —  sin2  6  is  >  =  <  o,  that  is  according  as 
tan  #<=:>  tan  v. 

2°.  For  the  cylinder  of  revolution  about  the  axis  of  z,  we  make 
x=x'  cos  0,y  —y'  in  its  equation  x*+y*  =  r*;  /,  the  curve  of  in 
tersection  is  a-'2  cos2  04-/2=  r*  an  ellipse. 

EXAMPLES   (Coordinates  Rectangular). 

1.  Find  the  right  line  generators  of  the  hyperboloid 

*+>_£:=, 

9         4         i 

for  the  point  (2,  3  ?)  on  the  surface. 

2.  Find  the  right  line  generators  of  the  paraboloid  4jy2—  2$z2=  icxxr 
for  the  point  (?  2.  i)  on  the  surface. 

3.  Find  the  planes  of  circular  sections  of  the  following  surfaces  : 

36       (0 
=i44  (2) 
i2       (3) 
(4) 

^  -f.  j;2  g2 

4.  In  the  hyperboloid  of  revolution  of  one  sheet  —  f  ---  r  =  i 

find  the   equations   of  the  generating  line  whose   projection  on  the 
plane  xz  is  tangent  to  hyperbolic  section  in  that  plane  at  its  vertex. 

5.  Find  the  sections  of  the  cone  x*+y*=(z—  2)'2  by  planes  con 
taining  the  axis  of  j/,  at  angles  to  the  plane  xy  of  30°,  45°,  and  60° 
respectively. 

6.  Find  the  curve  of  intersection  of  the  surface 


by   a   plane  inclined  at  an  angle  of  30°  to  the  plane  xy,  and  whose 
trace  on  that  plane  makes  an  angle  of  45°  with  the  axis  O.v. 


CHAPTER   X. 

TANGENT    PLANES,   DIAMETRAL  PLANES,  AND 
CONJUGATE  DIAMETERS. 

123.   Straight  line  meeting  surfaces  of  second  order. 
We  can  transform  the  general  equation 


to  polar  coordinates  by  writing  x  =  /r,  y  —  mrt  z  —  nr,  (when  /,  m,  n 
are  in  rectangular  coordinates,  direction  cosines,  and  in  oblique  co 
ordinates,  direction  ratios).  The  equation  becomes 


r*(  A/2  +  Bfl*2  +  Cri2  +  2K'?nn  +  2  Win  +  2C7m) 

+C"«   +  F  =  o. 


Hence  a  straight  line  meets  the  surface  in  two  points,  and  if  these 
two  points  be  coincident  the  line  is  tangent  to  the  surface. 

124.    Tangent  Plane  to  surfaces  of  second  order. 

Let  the  origin  be  on  the  surface  (and  therefore  F=  o)  then  one 
of  the  values  of  r  in  (2)  is  r  =  o.  Now,  in  order  that  the  radius 
vector  shall  touch  the  surface  at  the  origin,  the  second  root  must  be 
o,  and  the  condition  for  this  is  A"/+B"w  +  C"n  —  o.  Multiplying 
this  by  r  and  replacing  Ir,  mr,  nr  by  x,  j>,  z,  this  becomes 


=  o.      (3) 

Hence  the  radius  vector  touching  the  surface  at  the  origin  lies  in  the 
fixed  plane  (3);  and  as  /,  ;;/,  n  are  arbitrary,  A",r  +  B'j  +  C"2  =  c  is 
the  locus  of  all  the  radii  vectores  which  touch  the  surface  at  the 
origin,  and  is  therefore  the  tangent  plane  at  the  origin. 

Hence,  if  the   equation   of  the  surface  can  be  written  in  the  form 
«„  +  «!=  o  (where  u.2  represents  terms  of  second  degree  and  u{  terms 

76 


NOTES   ON   SOLID    GEOMETRY.  ^ 

of  first  degree  in  #,  y,  and  z),  then  ul=  o  is  the  equation  of  the  tan 
gent  plane  at  the  origin. 

Therefore,  to  find  the  equation  to  the  tangent  plane  to  the  surface  at  the 
point  x'y'z',  transfer  the  origin  to  this  point.  rl  he  equation  may  then  be 
writ/en  u2  -f  Uj  =  o,  and  Uj  =  o  is  the  tangent  plane  referred  to  the 
point  of  contact  as  origin  ;  then  in  Uj=  o  retransfer  the  origin  to  the 
primitive  one. 

125.   For  the   central   surfaces  (origin  at  centre)  take  the  equation 


and  let  (x',y',  z')  be  the  point  of  contact.     Transferring  the  origin  to 
the  point  (x',  yr,  z')  by  the  formulae 

X  =  X  +  X    \ 

y  —  y  +  y  V  we  have 

2=  Z  +  Z1} 

Ax*  +  B/  +  Os8  +  2  Axx'  +  2  fyy'  +  2Czz'=o. 
Hence  the  tangent  plane  at  the  new  origin  is 
Axx'  -f  By/  +  Czz  =  o.      (  i  ) 

Now  retransfer  the  origin  for  equation  (i)   to   the  centre  by  the 
formulae 

x  =  x  —  x'  } 

y  =  y  —y'  \  and  we  obtain 

*=*-*') 


+  Czz'—  Ax'*—  Bi/2-  CVa=  o, 

or  AAVT'  +  BJ/  +  Czz'=i  (2)   the   required   equation   of  the  tangent 
plane,  at  the  point  x'y'z'  referred  to  centre. 

i°.   For  the  sphere  A  =B  =  C  =  -^  . 


a 
Hence  (2)  gives  xx  +yy'  +  zz'=  a\    (3) 

2°.  For  the  ellipsoid    A  =  -i -,  B  =  -1  C=  i 
xx'     yy'      zz'  _ 

3°.   For  the  hyperboloid  of  two  sheets  A  =  ~  E~ ^  C  =  — \ 

xx        yy        zz' 

7* 


78  .VOTES  ON  SOLID   GEOMETRY. 

4°.  For  the  hyperboloid  of  one  sheet,  A=— ,  B=:— ,  C= ^. 


xx        yv 


126.  For  the  surfaces  which  have  no  centre  (origin  at  vertex)  by 
treating   the    equation    By2  +  Cz*  =  2A.'x  in   a   similar  manner   we 
obtain 

Bj/y'  +  Oss'  =  K'(x  +  x')  (7)  for  the  equation  to  the  tangent  plane 
to  the  elliptical  paraboloid  and 

Bij/— Czz  =  A.'(x  +  x')  (8)  for  the  tangent  plane  to  the  hyper 
bolic  paraboloid. 

Remark.  The  same  method  may  be  applied  to  cones  and  cy 
linders. 

127.  Polar  planes  to  surfaces  of  second  order.     The  equations  (3) 
(4)  (5)  (6)  (7)  (8)  are  the  equations  to  the  polar  planes  to  the  sur 
faces  respectively  with  respect  to  the  point  (.vf,  y',  z')  and  these  polar 
planes  possess  properties  analogous  to  the  polar  lines  to  the  conic 
sections. 

128.  The  length  of  the  perpendicular  from  the  centre  on  the  tangent 
plane  to  the  ellipsoid  is  p  =  \Az8  cos2  a  +  &*  cos2  ft  +  c*  cos'2  y  ,    when 
cos  a,  cos  (3,  cos  y  are  its  direction  cosines. 

The  equation  to  the  tangent  plane  is  "-^  +  "-jrjH — g-  =  i.    It  may 

also  be  written  x  cos  a  +y  cos  ft  +  z  cos  y  —  p.     Hence  we  must 
have 

/      cos  a        cos  /?       cos  y a  cos  a  _  b  cos  ft  _  c  cos  y  _ 

i  ~     x'  y'  z'  x'  y  _s/ 

~J  ^~  ~^~  ~a~  ~T  ~7 


Hence  calling  the  direction  cosines  /,  »z,  w,  the  equation  of  the 
tangent  plane  may  be  written 


Ix  4-  my  -f  «.s  =  V^'2  +  ^2^2  +  ^9»2  •      (  9  ) 


NOTES   ON  SOLID   GEOMETRY. 


79 


129.    To  find  the  condition  thai  the  plane 


3C  \f  2 

f-  -~  H =i      (i)  shall  be  tangent  to 

a         ft         y 


x*         /         z* 

the  ellipsoid  —5-  +  ~—-r  -\ r 

a  u  c 

xx'        yy'       22' 
Comparing  (i)  with    — ^  +  '-—^  -\ — - 


we  must  have 


I      _  X'          I      _  y  I 

—      "  —         o~  j          7T~    —       /o     i       ' 

a         «2        ft         6*        y 


a          x         b  y  c          z 

or  -  —  — ,    — r  =  -^-,       -  =  — ;  /.   squaring  and 

^f         a         ft          by          c 

adding  — r  +  — ^  H r  —  i    is  the   required  condi- 

<^  p          y 

tion. 

130.  The  sum  of  the  squares  of  the  perpendiculars  \>,  p',  \>" ,  from  the 
centre  of  the  ellipsoid  on  three  tangent  planes  mutually  at  right  angles  is 
constant. 

Let  cos  a  cos  ft  cos  y\  cos  <*'  cos  ft'  cos  j/,  etc.,  be  the  direction 
cosines. 

Then  p2  =  «2  cos2  «  -f  <$2  cos2  /5  +<;2  cos2  y 

p'*  =  a"  cos2  a'  +b~  cos2  /?'  +tf  cos2  y' 
p' 2  =  <z2  cos2  a"  +  ^2  cos2  ft"  •}- c1  cos2  T/', 

and  adding  we  have 


131.  Cor.  Hence  the  locus  of  the  point  of  intersection  of  three  tangent 
planes  to  the  ellipsoid  which  intersect  at  right  angles  is  a  concentric  sphere 
of  the  radius  V V  +  b*  +  c1. 

For  02  the  square  of  its  distance  from  the  centre  is  equal  to 
p*+p'*+p'"\  and  therefore  to  a*  +  tf  +  c\ 

Remark.  In  the  case  of  hyperboloids  one  at  least  of  the  quantities 
a2.  b~,  c2  is  negative,  and  hence  their  sum  may  be  negative  or  nothing  ; 
in  the  former  case  there  is  no  point  in  space  through  which  three 
rectangular  planes  touching  the  hyperboloid  can  be  drawn,  and  in 
the  latter  case  the  centre  is  the  only  point  which  has  that  property. 


8o  NOTES   OA'   SOLID    GEOMETRY. 

132.  Diametral  Planes.      Definition.     A   diametral  surface  is    the 
locus  of  the   middle  points   of  a  series  of  parallel  chords  of  a  given 
surface.      Diametral  lines  or  diameters  are  the  intersections   of  the 
diametral  surfaces. 

133.  To  find  the  diametral  surface  corresponding  to  a  given  series  of 
parallel  chords  in  a  surface  of  the  second  order  which  has  a  centre. 

Let  the  equation  of  the  surface  be 

/,  m,  n  the  direction  cosines  of  each  the  parallel  chords,  and  #',  /,  z' 
the  coordinates  of  its  middle  point. 
The  equation  of  the  chord  will  be 

x — oc         y — v         z — z 

m  —     — —  =:   —         —  zn  i*. 


I  m  n 

Then  for  the  points  in  which    it  meets  the  surface  (i)  we  shall 
have 


or 


Imposing  on  this  the  condition  of  equal  roots  for  r,  we  have 
Atx'  +  Bniy'  +  Cnz'  =  o  (2)  the  equation  of  the  diametral  surface,  a 
plane  passing  through  the  centre. 


134.   The  diameter  —  =  —  =  —    is  one  of  the   series   of  parallel 
/        m       n 

chords  bisected  by  the  plane  (2),  and  is  called  the  diameter  conju 
gate  to  the  plane,  and  conversely  the  plane  lx  +  my  +  nz  —  o  is  con- 

A.v       Bv       Cz 

luxate  to  the  diameter   -  —  =  —  =  —  -. 
I          m          n 

If  a  diametral  plane  be  chosen  as  a  new  plane  of  xy  and  its  con 
jugate  diameter  be  taken  as  the  new  axis  of  z,  the  centre  O  being  still 
the  origin;  then,  since  every  chord  parallel  to  Oz  is  bisected  by 
the  plane  xy,  the  equation  of  surface  will  contain  only  the  second 
power  of  z.  Hence,  if  there  be  three  planes  through  the  centre  the 
intersection  of  any  two  of  which  is  conjugate  to  the  third,  the  equa 
tion  of  the  surface  referred  to  these  planes  will  be  of  the  form 

AV  +  By  +  CV=i,     (3) 
that  is  of  the  same  form  as  the  equation  referred  to  rectangular  axes. 


NOTES  ON   SOLID    GEOMETRY.  8  1 

!35-  To  find  the  conditions  that  of  three  planes  through  the  centre  of  a 
surface  of  the  second  order  each  may  be  diametral  to  the  intersection  of  the 
other  two. 

Let  the  planes  be 

/  v  _|_  mv  +  nz  =  0,   I'x  +  my  +  n'z  =  o,  l"x  4-  m  '  !y  +  n"z  =  o. 
The  equations  of  the  diameters  conjugate  to  the  first  plane  are 


and  if  this  be  parallel  to  the  other  two  planes,  we  shall  have 
/          .  m         ,  n  .  ...  I         ,,  m        ,,  n 

/'A  +  M  B+"c=oandrA+'"  B+"  c=0; 

these  with  the  third  equation  I'  —  +  m   —  +  n'  —  —  o,    found    in 

like  manner,  are  the  required  conditions. 

These  three  planes  are  called  conjugate  planes,  and  their  intersec 
tions  conjugate  diameters. 

Since  we  have  only  three  relations  between  the  six  quantities  there 
will  be  an  infinite  number  of  systems  of  conjugate  planes  in  each 
surface. 

136.  Equations  referred  to  conjugate  diameters.  If  in  (3)  Art.  134  we 
make 

A>_  '    B'_JL  c-- 
_  a,v  i  -  b,v  ^  -  -& 

Then  for  the  ellipsoid 

X~        I'2        22 

—  +'—  -f—  =  i    will   be  the  equation  referred   to  conjugate  di 

ameters,  and  a,  b'  ,  c  will  be  the  semi-conjugate  diameters. 
For  the  hyperboloids  we  shall  have 

a-2     y     z*          ,  x1     y     z2 

^-^-7T=  i  and  jr^-^bc  r. 

Remark.  The  tangent  planes  at  the  extremities  (.*•',  j/,  z)  of  any 
diameter  to  a  central  surface  are  parallel  to  the  diametral  plane 
conjugate  to  the  diameter  so  that  the  conjugate  plane  of  the  diameter 
through  the  point  (x1,  y',  z'}  on  the  ellipsoid  is 

xx'       vy'      zz' 


82  NOTES  ON  SOLID    GEOMETRY. 

137.  The  sum  of  the  squares  of  three  conjugate  semi-diameters  of  the 
ellipsoid  is  constant. 

In  the  first  place,  any  point  on  the  ellipsoid  may  be  represented  by 
the  equations  x  =.  a  cos  A,  y  =  b  cos  //,  z  =  c  cos  Y,  when  cos  A, 
cos  ju,  cos  v  are  the  direction  cosines  of  some  line,  for  the  condition 
cos2  A  +  cos2  //-fcos2  v  —  i  cause  these  three  equations  to  satisfy  the 
equation  of  the  ellipsoid. 

Therefore  if  cos  A,  cos  fa  cos  r,  cos  A',  cos  //,  cos  v'  are  the  direc 
tion  cosines  of  two  lines  answering  to  the  extremities  of  two  conju 
gate  diameters,  these  will  be  at  right  angles  to  each  other. 

T,  xx        yy'     zz 

ror  the  equation  — —  +~j^-  +  — g-=  o  will  give 

cos  A  cos  A'  +  cos  yu  cos  //4-cos  v  cos  v   =  o. 

Now  the  square  of  the  length  of  any  semi-diameter  ~v'2+y2+V2 
expressed  in  terms  of  A,  /i,  v,  is 

rt'2=  a*  cos2  A  +  <£2  cos2  JJL  +  C*  cos2  r, 
and  of  the  conjugates  in  terms  of  A',  //,  v\  A",  //',  y" 
£'2=rt2  cos2  A'+£2  cos2  yu'  +  r  cos2  r' 
<:"*=  a*  cos2  A"+32  cos2  //'  +  r>  cos2  v". 
Adding  we  have 

a"2  +  3/2  +  c'2  =  a2  +  tf  +  <r2,  since  the  lines  A,  ju,  r,  A',  //',  rx,  and 
A",  yu",  y"  are  mutually  at  right  angles. 

138.  To  find  the  locus  of  the  intersection  of  three  tangent  planes  at  the 
extremities  of  three  conjugate  diameters. 

The  equations  of  the  three  tangent  planes  are 

—  cos  A  +  v  cos  u,  -f  -  cos  v   —  i 
a  o  c 

OC  V  Z 

—  cos  A'  +  —  cos  v  -\-  -  cos  v'  =  i 

(2  O  C 

x  y          ,,       z 

-  COS  A     -f  '—  COS  V      -\ COS  V      —    I . 

a  b  c 


Squaring  and  adding,  we  get  for  the  equation  of  the  locus 
>H 

tV~T- 


~-f--r=  3  an  ellipsoid  with  the  semi-axes  a\/~$~,  b^/  3  , 


NOTES   ON  SOLID   GEOMETRY.  83 

139.  The  parallelepiped  whose  edges  are  three  conjugate  semi-diameters 
of  an  ellipsoid  lias  a  constant  volume. 

Let  Qx,  Oy,  O.2  be  the  semi-axes  of  the  surface  a,  b,  c\  Oo;',  Oy, 
Ozr  any  system  of  semi-conjugate  diameters  a  ',  b'  ,  c\  let  the  plane 
of  x'y'  intersect  that  of  xy  in  the  semi-diameter  OjCi—  A,  and  let 
Oy2=  B  be  the  semi-diameter  of  the  curve  ^  #'  which  is  conjugate  to 
CXrj.  Hence  parallelogram  a'b'=  parallelogram  AB. 

.-.   Vol  (a,  b\  f')  =  Vol  (A,  B,-O 

for  these  figures  have  the  same  altitudes  and  equal  bases. 

Let  the  plane  z'Oj'2  intersect  xy  in  the  semi-diameter  Oi^  —  C, 
then  this  plane  must  contain  Oz  ;  for,  being  conjugate  to  OA*,  in  a 
principal  plane  it  must  be  perpendicular  to  that  plane  ;  hence  CXv,, 
Or,.  Oz  form  a  system  of  semi-conjugate  diameters,  and  any  two  of 
them  are  semi-conjugate  diameters  of  the  plane  section  in  which  they 
are  situated. 

.-.   Vol  (A,  B,  c')  =Vol  (A,  C,  c) 

Vol  (A,  C,  c)  =Vol  (a,  b,-c) 
.'.   Vol  (a1,  b',  c')  =Vol  (a,  6,  c). 

140.    To  find  the  diametral  plane  bisecting  a  given  system  of  parallel 
chords  in  the  case  of  the  surfaces  which  have  not  a  centre. 
Taking  the  equation  of  the  surface 


r  ,       ,       ,     x—x        v—y'       z  —  z' 
and  one  or  the  chords  -       —  =  —      -  =.  ~    -  =  r 
I  m  n 

the  equation  of  the  diametral  plane  will  be 

;;/  BJ/  +  W  Cz=  A'7. 

Hence  the  diametral  planes  are  parallel  to  the  common  axis  of  the 
principal  parabolic  sections. 

We  cannot,  therefore,  in  these  surfaces  have  a  system  of  three  con 
jugate  planes  at  a  finite  distance,  but  we  can  find  an  infinite  number 
such  that  for  two  of  them  each  bisects  the  chords  parallel  to  the  other 
and  to  a  third  plane,  by  proceeding  as  in  Art.  (135). 

By  taking  the  origin  where  the  intersection  of  these  two  meets  the 
paraboloid,  and  referring  to  these  three  planes,  the  equation  of  the 
surface  will  be  of  the  form 


84  NOTES   OiV   SOLID   GEOMETRY. 

And  the  third  plane  is  evidently  the  tangent  plane   to  the  surface  at 
the  new  origin. 

141.  The  tangent  planes  to  the  hyperboloid  of  one  sheet  and  /he  hyperbolic 
paraboloid  at  a  point  x'y'z'  intersect  the  surfaces  each  in  two  right  line 
generators  through  the  point  of  contact. 

The  equation  of  the  hyperboloid  of  one  sheet  referred  to  any  con 
jugate  diameters  is 


and  the  equation  of  the  section  made  by  any  plane  y  =  ft  parallel 
to  the  conjugate  plane  of  xz,  is 


and   it  is  evident  that  the  value  /3—b'  gives  us  the  section  of  the 
tangent  plane  at  the   extremity  (.v',  y' ,  z')   of  the  diameter  b';   or 

x3         z"* 

~~^ 7?  —  °>    two  right  line  generators. 

Ct  c 

For  the  hyperbolic  paraboloid 

B>8-CV=2E".,v    (i) 

the  tangent  plane  through  the   origin  is  x  =  o,  and  its  intersection 
with  (i)  is 

B^'2— CV=  o,  two  right  line  generators. 


CHAPTER  XL 

GENERAL    EQUATION    OF    THE    SECOND    DEGREE    IN 

x,y,  AND  z. 

142.   In  order  to  discover  all  the  surfaces  represented  by  the  gen 
eral  numerical  equation 


Bj/2 

Gz2  +2C'^  +  2C"a  =  D  (E). 

we  will  first  transform  the  coordinates  to  a  new  origin  by  means  of 
the  formulae 

x  •=.  a  +  x*  \ 


z  — 

and  endeavor  to  determine  the  coordinates  (a,  /?,  y)  of  the  new 
origin  in  such  manner  as  to  cause  the  terms  of  the  first  degree  to  dis 
appear.  If  this  can  be  effected  the  equation  will  be  reduced  to  the 
form 

Ax2  4-  Br  +  Cs2  +  2A.'zy  +  2%'zx  +  2C'xy  =  F'     (F) 

in  which  there  is  no  change  when  —  x,  —y,  —z  are  substituted  for 
+  x,  +y,  +z,  and  which  therefore  represents  a  surface  having  a 
centre,  and  the  new  origin  of  coordinates  is  at  this  centre. 

Now,  several  different  cases  may  arise  according  to  the  numerical 
relations  among  the  coefficients  A,  B,  C,  Ax,  B',  C',  A",  B",  C". 

i°.  a,  ft,  y  the  coordinates  of  the  centre  may  each  have  a  finite 
value  found  from  the  three  equations  determining  the  conditions  of 
the  transformation. 

2°.  a,  /3,  y  may  have  infinite  values. 

3°.  a,  /?,  y  may  be  indeterminate. 

8  85 


86  AT07^ES   ON   SOLID   GEOMETRY. 

The  surfaces  corresponding  to  these  three  cases  will  be 

(A)  Surfaces  having  a  centre. 

(B)  Surfaces  having  no  centre  (centre  at  an  infinite  distance). 

(C)  Surfaces  having  an  indefinite  number  of  centres. 

143.   Making  the  actual  transformation  of  (E)  by  the  formulae  (i 
we  have 


And  in  order  that  the  terms  of  the  first  degree  in  x,  y,  z  shall  dis 
appear,  we  must  have 


(C) 


which  are  called  the  equations  of  the  centre. 

1°.  If  these  three  equations  give  finite  values  for  a.  /?,  y,  then 
the  surface  represented  by  the  given  equation  has  a  centre. 

2°.  If  two  of  ihese  equations  are  incompatible  this  shows  infinite 
values  for  a,  /3,  y,  and  the  surface  has  no  centre. 

3°.  If  the  three  equations  reduce  to  two,  then  the  surface  has  a 
line  of  centres.  For  each  one  of  the  equations  is  the  equation  of  a 
plane,  and  two  taken  simultaneously  represent  a  line,  and  the  surface 
is  an  elliptical  or  hyperbolic  cylinder.  For,  cut  the  surface  by  the 
planes  P  and  Q,  P  cutting  the  line  of  centres  (D)  and  Q  containing 
that  line.  The  section  by  P  is  a  curve  of  the  second  degree  having 
its  centre  on  the  line  D,  and  hence  an  ellipse  or  hyperbola.  The 
section  Q  will  be  two  straight  lines  parallel  to  the  line  D,  and  as  Q 
may  revolve  about  D  in  all  its  positions  giving  two  straight  line  sec 
tions  parallel  to  D,  the  surface  is  a  cylinder. 

4°.  If  the  three  equations  reduce  to  a  single  one,  then  the  surface 
has  a  plane  of  centres  (i.  e.,  the  given  equation  represents  coincident 
or  parallel  planes). 

Note.     The  equations  of  the  centre  can   be  found   in   any  given 


NOTES   ON   SOLID    GEOMETRY.  87 

equation  most  readily  by  finding  the  derived  equations  with  regard 
to  x,  y,  and  z  respectively  (i.e.,  by  differentiating  with  regard  to  x.  y,  z 
respectively),  the  x,  j',  and  2  in  the  resulting  equations  standing  for 
a,  ft,  y. 

144.   Example  i.     Determine  the  class  of  the  surface  represented 
by  the  equation  ,v2  +  ^y1  +  422  +  2yz  +  \zx  +  6.17  —  2  6x  —  2$y — 322=  26. 

The  equations  of  the  centre  are 

6>'  +  22  +  6x—  24  =  O  >  .    These  give  y  =  2  I 
82  +  27  +  4^—32  =  0)  *=3J 

and  the  surface  has  a  centre. 

Example  2.    Determine  the  class  of  the  surface 

The  equations  of  the  centre  are 

2JI-  +  22+2  V  — 4  =  O  ) 

f 

2 y  +  22  +  2X—  2  =  O   ,x- 

—  4  Z  +  2y  +  2X  +  2  =  O  ) 

the  first  two  of  which  x+y+z  =  2,  x+y  +  z  =  I  are  incompatible, 
hence  the  coordinates  of  the  centre  are  infinite,  and  the  surface  has 
no  centre. 

Example  3.   Determine  the  class  of  the  surface 

— 22 — 2yz — zx-\-^xy+2z  =  o. 
The  equations  of  the  centre  are 


2x — z   +4y  ~  o 
8j/ —  22  +  4^v  =  o 

—  2Z  —  2V  —  X+  2=OJ 

The  first  two  of  these  are  identical,  hence  the  three  equations  re 
duce  to  two  and  the  surface  has  a  line  of  centres  (/.  e.}  is  a  cylinder). 
Example  4.    Determine  the  class  of  the  surface 

8-v2  +  1 8v2  +  222  +  1 272  +  8zx  +  2^xy —  5o.v — 757 —  252  +  75  =  o. 
The  equations  of  the  centre  are 

1 6x  +   82  +  247—  50  =  o 


= 
i2r+    8.v— 25  =  o 


88 


NOTES   OX  SOLID   GEOMETRY. 


which  are  all  three  the  same,  each  being  8.r+  i2>' 4-4.3  =25.  Hence 
the  surface  has  a  plane  of  centres,  and  consists  of  a  pair  of  parallel 
planes. 

145.  Recurring  to  the  general  equations  of  the  centre 
A  <*4-C7?  +  By  +  A"=  o  ] 

•  +  ir=o(  (c) 


we  may  find  an  easy  rule  for  a  relation  among  the  coefficients  in  any 
given  equation  by  which  we  can  distinguish  the  central  surfaces  from 
those  having  no  centre  and  those  having  an  infinity  of  centres. 

The  common  denominator  of  the  values  of  a,  /3,  and  y  in  these 
equations  is  the  determinant 


A,  C',  B' 
C',  B,  A' 
B',  A',  C 


'2  +  CC'2-ABC-2A'B'C'. 


Now,  if  R  be  different  from   zero,  the  surface  has  a  centre  ;  but  if 
R  =  o  it  may  either  have  no  centre  or  an  infinity  of  centres. 

The  value   of  R  may  be  written  out   by  the  following   mnemonic 
A,  B,  C 


form  : 


A'  B'  C' 

A'  B'  C' 


the  letters  to  be  multiplied  by  columns  for  the  first 
three  terms,  and  by  rows  for  the  two  last. 


146.  To  find  an  easy  rule  for  F,  the  new  absolute  term  in  the  trans 
formed  equation  of  the  central  surfaces  when  the  origin  is  moved  to  the 
centre. 

This  complete  transformed  equation  is 

A_vs  +  By2  +  Or  +  2  A'zy  +  zR'zx  +  2C'xy  =  F     when 


F=D- 


Now,  multiplying  the   first  of  the  equations  (C)  of  the  centre  by 
a,  the  second  by  ft,  and  the  third  by  y,  and  adding  them 


we  have 


NOTES  ON   SOLID    GEOMETRY. 


89 


Hence  F  =  D  —  (A"« +  B"/?  +  C'».  Therefore  the  rule  for  F  is 
substitute  for  x,  y,  z  in  the  terms  of  the  first  degree  one-half  the  co 
ordinates  of  the  centre  (i.  e.,  ^a,  |-/?,  ^y  respectively  and  take  result 
from  D. 

Example  i.    Taking  the  Example  i,  Art.  (144), in  which  the  coordi 
nates  of  ihe  centre  are  found  to  be  x  =-i,y  —  2,  z  =  3, 
we  have  F=26  +  26  x  J-  +  24  x  i  +32  x  f  —  1 1 1; 

and  the  transformed  equation  is 

X*  +  3>'2  +  432  +  2yz  +  4zx  -f  6,r>'  =  1 1 1 . 

Ex.  2.      2^ +  3y*  +  4Zz  +  Syz  +  6xz  +  4xv— 6x — Sy—I4z  =  20. 
Here  the  coordinates  of  the  centre  are  x  =  \>y  =  2,  s=  — i. 

.'.  F==2o  +  6  x-J  +  8  x  i  +  14  x  — 1  =  17; 
and  the  transformed  equation  is 

2  A*  4-  3/  +  4-s2  +  870  +  6^2-  +  ^xy  —  1 7. 

147.  Removal  of  the  terms  in  xy,  xz,  yz.    Reduction  of  the  equation  of 
the  second  degree  to  two  forms. 

Fora  more  complete  discrimination  of  the  surfaces  represented  by 
the  general  equation,  we  will  now  remove  the  terms  in^ry,  xz,  yz  by 
a  transformation  of  coordinates.  So  far  we  have  made  no  supposition 
as  to  the  direction  of  the  axes.  Henceforth,  for  convenience,  we  will 
consider  the  axes  rectangular. 

Taking  the  equation  (E)  in  rectangular  axes  we  propose  now  to 
transform  it  to  a  system  also  rectangular  in  such  manner  that  the 
terms  in  xy,  xz,  yz  shall  disappear.  The  disappearance  of  these 
terms  can  only  be  effected  by  taking  for  coordinate  planes  either  dia 
metral  planes  or  planes  parallel  to  them. 

We  will  therefore  begin  by  finding  a  diametral  plane  conjugate  to 
a  given  diameter. 

148.  To  find  a  diametral  plane  conjugate  to  a  given  diameter. 

x — x' y—y'  __  z—z' 

I  m  n 

Putting  x-=  x'  +  lr,  y  —  y'  +  mr,  z=z'  -\-nr  in  the  general  equation, 
and  arranging  with  reference  to  r,  we  have  for  the  coefficient  of  the 
first  deree  in  r 


QO  NOTES   ON  SOLID   GEOMETRY, 

and  this  placed  equal  to  zero  is  the  equation  of  the  diametral  plane, 
namely 

( A/+  B'»  +  C'm)x  +  (C7+  Bm  +  A'n}v  +  (B7+  A'm  +  Cn)z  +  A'7 


149.    70  determine  a  diametral  plane  perpendicular  to  the  chords  which 
it  bisects,  that  is,  to  find  a  principal  plane. 

In  order  that  the  diametral  plane  shall  be  perpendicular  to  the  line 

x  —  a          v  —  b         z  —  c 


m  n 


,  we  must  have  the  conditions  fulfilled 


I  m 

or  putting  each  of  these  equal  to  s. 

At  +R'n+C'm  = 


(A) 


and  also  the  condition  /2  +  w2  -f  #2  =  i  . 

To  determine  /,  w,  and  n  in  equations  (A)  we  first  find  s.      Writ 
ing  these  equations 


they  give  the  result 


A-J,  C',  B' 
C',  B  — j,  A' 
B',  A',  C-.r 


=  o 


or 


or 


C'*-ABC-2A'B'Cf=o     (D). 

This  cubic  has  necessarily  one  real  value  for  s,  which  substituted 
in  (A)  gives  one  set  of  real  values  for  /,  m,  n.  Hence  there  is  one 
principal  plane. 

For  convenience  of  discussion  let  us  take  this  plane  perpendicular 


NOTES  ON   SOLID   GEOMETRY.  gi 

to  the  axis  of  z,  then  /  =  o,  w  =  o,  and  n  =i.  And  hence  equa 
tions  (A)  give  B'=  o,  A'=  o,  and  the  general  equation  transformed 
to  this  principal  plane'as  plane  of  xy  is  of  the  form 


z  =  D. 

Now  we  know  from  the  like  discussions  in  conic  sections  that  one 
transformation  is  always  possible,  and  but  one  to  a  system  of  rectan 
gular  axes  in  the  plane  xy  which  shall  cause  the  term  in  xy  to  dis 
appear.  Hence  there  are  three  principal  planes,  and  three  sets  of 
values  for  /,  m,  n,  and  the  cubic  (D)  has  three  real  roots. 

The  general  equation  may  then  be  always  reduced  in  rectangular 
coordinates  to  the  form 


'*  =  D.    (E') 
which  represents  then  all  the  surfaces  of  the  second  order. 

1  50.  The  reduction  of  this  equation  Lx2  +  My2  -f  Nz2  +  2L'x  +  2M'y 
+  2N  z  =  D  to  two  forms. 

i°.  If  L,  M,  and  N  are  different  from  o. 

Then  we  may  cause  the  terms   of  the  first  degree  to  disappear  by 

L'  M'  N' 

transferring  the  origin  to  the  point  x  =  —  ~r~>-J'  =  —  ivf  '  *      —  \f 

1  4  i  *  L  IN 

The  surface  will  then  have  (#',.>'',  z')  for  its  centre,  and  the  equation 
will  be  of  the  form 

Lr'  +  My'  +  Ns2^  F.     (I.) 

2°.  If  one  of  the  three  coefficients,  L,  M,  N,  for  example  L  =  o 
and  L'  be  different  from  o. 

We  cannot  then  cause  the  term  2~L!x  to  disappear,  but  by  trans 
ferring  the  origin  to  the  point 

D'  M'  N' 

x'=  -—-.,  y  =  —  —  —   z'=  --  ^p-   the  equation  will   take  the  form 
2L  M  N 

M/  +  Nsf=  2V  x.     (II.)* 

The  forms  I.  and  II.,  we  have  seen,  belong  to  the  surfaces  of  the 
second  order,  which  we  have  already  discussed.  Hence  the  general 
equation  of  the  second  degree  (E)  represents  these  surfaces  and  no 
others. 


QO  NOTES  ON  SOLID   GEOMETRY. 

151.  The  form  I.  we   have  seen  represents   the  ellipsoid,  the  two 
hyperboloids  and  cones  of  second  degree,  and  includes  the  elliptic 
and  hyperbolic  cylinder,  Mv2  +  Ns2  -—  F  and  parallel  planes  N22=F. 

The  form  II.  represents  the  elliptic  and  hyperbolic  paraboloids, 
and  the  parabolic  cylinder. 

152.  The  complete  reduction  of  the  equation  of  the  second  degree  to  the 
simple  forms  I.  and  II.      Use  of  the  discriminating  cubic  (D). 

The  resolution  of  the  equations  (A)  furnishes  for  each  value  of  s 
in  the  cubic  (D),  one  system  of  values  of/,  m,  n.  We  have  then  three 
systems,  /,  m,  n]  /',  ??i',  n  '•  /",  m",  n  ',  which  are  the  direction  cosines 
of  the  three  rectangular  axes  {principal  axes']  to  which  the  surface 
must  be  referred  in  order  to  cause  the  products  .%T,  xz,  yz  to  dis 
appear  ;  the  formulae  of  transformation  are  then 


y  —  mx  +  my  +  m"z' 
z  =  nx'  +  n'y'  +  n"z'. 

If  we  take  only  the  terms  in  x*  in  this  substitution  we  find 
L  =  A/2  +  Bw2  +  C;*2  +  2h.'mn  +  2BW+  zC'lm. 

But  if  we  multiply  the  equations  (A)  respectively  by  /,  m,  n  and  add, 
remembering  that  /2  +  7«2  +  7z2  =  i  we  have 

A/2  +  B;;;2  +  CV  +  2h.'mn  +  2  BW+  2C'lm  —  s  ; 

Hence  L  is  a  root  of  the  cubic  (D)  and  M  and  N  are  the  other  two 
roots. 

For  the  values  of  L',  M',  N'  we  will  have 

L'  =  A'7   +B"w  +C"n     \ 
M'=  A  '7  +B";;/r  +  C'V          (M). 


The  absolute  term  D  does  not  change  in  this  transformation  since 
the  origin  is  not  changed  thereby. 

For  the  surfaces  having  a  single  centre  after  solving  the  cubic,  we 
have  only  to  calculate  F,  for  which  we  have  given  a  rule. 

For  the  surfaces  having  no  centre  the  coefficient  designated  by  V 
is  equal  to  —  L',  and  is  computed  by  first  finding  in  equations  (A) 
the  values  of/,  m,  n,  which  correspond  to  s  =  o.  Both  in  the  cases 
of  surfaces  having  no  centre  and  a  line  of  centres,  one  root  of  cubic 
=  o  and  we  have  only  a  quadratic  to  solve  to  determine  L  and  M. 


NOTES  ON  SOLID   GEOMETRY.  93 

153.  For  surfaces  having  a  centre,  if  \ve  wish  only  to  discover  the 
particular  class  of  the  surface,  without  making  the  complete  trans 
formation  of  the  equation  to  its  centre  and  axis,  the  sign  of  the  roots 
of  the  discriminating  cubic  will  tell  us  whether  the  surface  is  an  ellip 
soid,  hyperboloid  of  one  sheet,  or  hyperboloid  of  two  sheets.  These 
signs  we  can  ascertain  from  inspection  by  Descartes's  rule  *  without 
solving  the  equation. 

Example.  Find  the  nature  of  the  surface  7#2  +  6y2+5.s2— 4_>'0— $xy 
=  6.  The  cubic  (D)  gives 

s3—  (7  +  6  +  5)sa  +  (42  +  35  +  30 -4  —  4)5+28  +  20— 210=0;  or 
s3—i8s*  +  99^—162=0. 

.-.  The  row  of  signs  is  H 1 ,  three  changes  of  sign.  Hence 

all  the  roots  are  +  and  the  surface  is  an  ellipsoid. 

So  also  for  surfaces  having  a  line  of  centres,  the  signs  of  the  roots 
of  the  quadratic  into  which  the  discriminating  cubic  degenerates, 
serve  to  distinguish  the  elliptic  from  the  hyperbolic  cylinder. 

And  for  surfaces  having  no  centre,  the  signs  of  the  roots  distinguish 
the  elliptic  paraboloid  from  the  hyperbolic  paraboloid. 

154.  Recapitulation  of  the  method  of  reduction  of  numerical  equations 
of  the  second  degree  and  of  distinguishing  the  surfaces  represented  by  them. 

We  now  propose  to  give  the  mode  of  distinguishing  the  nature  of 
the  surface  represented  by  any  given  numerical  equation  of  the  second 
degree  in  x,  y,  and  z,  and  of  finding  its  principal  elements. 

I.    Form  the  equations  of  the  centre,  and  also  the  discriminating 
cubic  from  the  remembered  form 
s*-  (A  +  B  +  C)sz  +  ( AB  +  AC  +  BC  -  A'2-  B'2-C'2).r  +  AA'2  +  BB'2 

+  CC'2-ABC-2A'B'C'=  o, 

observing  that  the  absolute   term  is  equal  to  R,    the  denominator 
of  the  values  of  the  coordinates  of  the  centre  in  the  general  equation, 

ABC 
and  therefore  can  be  formed  by  the  mnemonic  A'B'C'     (Art.  145). 

A'B'C1 

Then 

155.  i°.    If  R  be  different  from  o,  the  surface  has  a  centre.     Find 


Note.  "All  the  roots  being  real  the  number  of  positive  roots  is  equal  to  the  number  of 
changes  of  sign  in  the  row  of  signs  of  the  terms,  and  the  number  of  negative  roots  is  equal  to  the 
number  of  continuations  of  sign." 


94 


NOTES  ON   SOLID   GEOMETRY. 


the  coordinates   of  the  centre  and  transform  to  the  centre  by  the  rule 
in  Art.  (146).     Determine  the  signs  of  the  roots  of  the  cubic  by  Des- 
cartes's  rule.     Then  calling  these  roots  L,  M,  and  N,  and  calling  F 
the  new  absolute  term  on  the  second  side  of  the  equation. 
Then 

a.  If  L,  M,  N  all  have  the  same  sign  as  F,  the  surface  is  an  ellip 
soid. 

b.  If  L,  M,  N  all  have  a  different  sign  from  F,  the  surface  is  im 
aginary. 

c.  If  two  only  of  the  roots  L,  M,  N  have  the  same  sign  as  F,  the 
surface  is  the  hyperboloid  of  one  sheet. 

d.  If  only  one  of  the  roots  L,  M,  N  has  the  same  sign  as  F,  the 
surface  is  the  hyperboloid  of  two  sheets. 

e.  If  F  =  o  and  L,  M,  N  all  have  the  same  sign,  the  locus  is  a 
point. 

f.  If  F  =  o  and   one  of  the  roots  L,  M,  N  has  a  different  sign 
from  the  other  two,  the  surface  is  an  elliptic  cone  (Art.  85). 

156.  2°.   If  R  =  o  the   cubic  has  one  of  its  roots  s  =  o  and  is 
degraded  to  a  quadratic,  the  coefficient  of  s,  namely  AB  +  AC  +  BC 
— A'2— B'2  — C'2,  becomes  the  absolute  term. 

And  if  the  equations  of  the  centre  are  incompatible  the  surface  has 
no  centre. 
Then 

a.  If  the  roots  M  and  N  of  the  quadratic  (degenerate  cubic)  have 
the  same  sign  (i.  e.)  if  AB  +  AC  +  BC  — A'2  — B'*—C'2>o  the  surface 
is  the  elliptical  paraboloid. 

b.  If  M  and  N  have  different  signs   (/.  e.)  if  AB  +  AC  +  BC-A" 

—  B'2—  C'*<o  the  surface  is  the  hyperbolic  paraboloid. 

c.  If  one  of  the  roots  M  or  N  be  zero  (*'.  e.)  if  AB  + AC  +  BC-A'2 

—  B'2 — C'2=  o  the  surface  is  the  parabolic  cylinder. 

157.  3°.    If  R  =  o  and  the  equations  of  the  centre  can  be  reduced 
to  two  equations,  the  surface  has  a  line  of  centres.     The  cubic  as  in 
(2°)  has  one  of  its  roots  S  =  o  and  degenerates  into  the  quadratic 

j«_(A+B  +  C)s  + AB  +  AC  +  BC --A"-B'»-C"  =  o. 

Then 

a.  If  the  roots  M  and  N  of  this  quadratic  have  the  same  sign  (i.  e.) 


NOTES  ON  SOLID  GEOMETRY. 


95 


if  AB  +  AC  +  BC  — A"'— B'2— C'~>  o  the  surface   is  an  elliptic  cy 
linder. 

b.  If  the  roots  M  and  N  have  different  signs  (i.  e.)  if  AB  +  AC 
+  BC— A'"2  —  B'2  —  C'2<  o  the  surface  is  the  hyperbolic  cylinder. 

c.  If  in  the  reduced  equation  of  the  cylinder  Mz2-f  Ny9  =  H,  H  be 
equal  to  o,  and  M  and  N  both  of  same  sign,  the  locus  is  a  straight 

z  =  o  } 
line  >  . 

J  =  o  \ 

d.  If  H  =  o  and  M  and  N  be  of  different  signs  the  surface  con 
sists  of  intersecting  planes. 

158.  4°.  If  R  =  o  and   the   equations  of  the    centre   become  a 
single  equation,  the  surface  has  a  plane  of  centres,  and  consists  of 
two  parallel  or  coincident  planes,  which  are  readily  found  by  solving 
the  equation  with  reference  to  any  one  of  the  variables. 

159.  5°.    In  the  case  of  surfaces  of  revolution  the  cubic  has  equal 
roots.     To  examine  the  cubic  for  equal  roots  in  the  case  of  central 
surfaces  of  revolution,  we  simply  look  for  a  commoe  root  between  it 
and  its  first  derived  equation  (differential). 

1 60.  GENERAL  REMARK.     In  any  of  the  above  cases  we  may  com 
plete  the  reduction  by  solving  the  cubic  to  get  the  new  axes  and 
thus  obtain   their  direction  by  finding  /,  m,  n  from  equations  (A). 
And  in  the  case  of  the  surfaces  without  a  centre  we  may  find  V, 
from  equations  (M). 

161.  REMARK  I.      In  the  cases  of  surfaces  having  a  line  of  centres 
and  of  those  not  having  a  centre,  we  can  distinguish  readily  the  sur 
face  represented  by  a  given  numerical  equation  through  sections  by 
the  coordinate  planes. 

i°.  If  the  equations  of  the  centre  show  a  line  of  centres,  sections 
by  the  coordinate  planes  will  tell  whether  the  surface  is  an  elliptic  or 
a  hyperbolic  cylinder. 

2°.   When  the  equations  of  the  centre  show  no  centre,  then 

a.  If  there   are  ellipses  among  these  sections  by  the  coordinate 
planes,  the  surface  is  an  elliptical  paraboloid. 

b.  If  there  are  hyperbolas  among   these  sections,  the  surface  is  a 
hyperbolic  paraboloid. 

c.  If  all   these   sections  are   parabolas,  or   one   of  them   parallel 
straight  lines,  the  surface  is  a  paraboMc  cylinder. 


96  NOTES   ON  SOLID   GEOMETRY. 

162.  REMARK  II.     Again,  if  the  terms  of  the  second  degree  in  the 
given  equation  break  up  into  unequal  real  factors,  the  surface  must 
be  either  the  hyperbolic  paraboloid  or  hyperbolic  cylinder,  and  these 
two  surfaces  are  otherwise  readily  distinguished.     We  may  note  also 
that  if  the  terms   of  the  second  degree  in  the  given  equation  form  a 
perfect   square,  the  surface    is    either   a    parabolic   cylinder   or  two 
parallel  planes. 

163.  We  will  now  illustrate  by  a  few  examples  : 

Ex.  i.      yjt-2—  iy~  +  6zi  +  2^xy+  \2yz-  1202:  =  ±84. 


As  this  is  a  central  surface  with  the  origin  at  the  centre,  we  only 
need  the  discriminating  cubic,  which  is 

j8  —  343^4-2058  =  o  ;  or  s*±os*—  343^4-2058  =  o. 

The  signs  4-  ±  --  1-  show  one  continuation  and  two  changes,  and 
hence  the  surface  is  a  hyperboloid  of  one  sheet,  or  two  sheets,  accord 
ing  to  the  sign  of  84. 

By  trial  we  find  that  7  is  a  root  of  the  cubic,  and  then  by  depress 
ing  the  equation  we  find  the  other  two  roots  are  14  and  —  21.  There 
fore  the  equation  of  the  surface  referred  to  its  centre  and  axes  is 

7*'-f  14^-212*=  ±84;  or.r2  +  2_y2-3S2=:±i2. 

Ex.  2.       2  5.r2  +  2  2_/  +  i6s24-  i6yz—4zx—  2oxy—  z6x—  40^—442 

=  -46. 

The  equations  of  the  centre  are 

250,'—  ioy—    22  =  13 
—  I  o:r  4-  2  2j  4-    82=20 
-   2X+    8y+  162  =  22  ; 

whence  we  find  the  coordinates  of  the  centre  x  —  \,y  =i,  *  =i. 
Moreover 

F  =  —  46  4-  26  •  \  +  40  •  J  +  44  •  I  =  9- 
The  discriminating  cubic  is 

/—  63^+  1  134^—5832  =  o. 

Its  signs  give  three  changes.  Hence  all  the  roots  are  positive. 
The  surface  then  is  an  ellipsoid. 


NOTES   ON   SOLID    GEOMETRY.  97 

By  trial  we  find  that  9  is  one  of  the  roots  of  the  cubic.     Hence  the 
other  two  are  18  and  36.     The  reduced  equation  is  then 

9jrJ-fi8>'2-f  36s2  —  9  ;  or  A-8 
And  the  principal  semi-axes  are  i    -  ••—  ,  __ 

V2        2 


Ex.  3.  5  x*  +  i  of  +  i  ys2  +  2  6yz  +  1  8zx  +  i  $xy  +  6x  +  ?>'  +  1  02  =  64. 
The  equations  of  the  centre  are 

5.r+    7^+   92  =-3 
7A:+iqy+  132  =—4 


Multiplying  the  first  of  these  equations  by  —  i  and  the  second 
by  2,  and  adding,  we  obtain  the  third.  Hence  the  equations  are 
only  two  independent  ones.  The  surface  is  therefore  a  cylinder.  In 
tersecting  it  by  the  coordinate  plane  xy,  i.  e.,  making  z  =  o,  we  obtain 

5^2  +  1  4xy  +  i  oy*  +  6x  +  Sy  —  •  64, 

which  is  an  ellipse.     The  surface  is  therefore  an  elliptic  cylinder. 
To  complete  the  reduction   we  transfer  the   origin   to   the  point 

where   the  line  of  centres  >  pierces  the  plane 


x,  y,  that  is,  to  the  point  z  —  o,  y  —  i,  x  —  —  2, 
and  find  F=64  +6—  4=:66. 

Also  the  discriminating  cubic  is 

.r3—  32,r-f6>9  =  o,   which  gives  j8  —  32^  +  6  =  o, 

the  roots  of  which  are  16  +  5^  10  and  16  —  5\/  10. 
And  the  reduced  equation  of  the  cylinder  is 


io)jr  +  (i6  —  5V7  io)>'2=66. 
Ex.  4  .      5  jp2  +  5/J  +  8s2  +  4sv  +  \zx  —  Sxy  +  6x  +  6y  —  30  =  o. 
The  equations  of  the  centre  are 


^4-4*= 


98  NOTES   ON   SOLID    GEOMETRY. 

Adding  the  two  last  of  these  equations  we  have 
$x—4y+2z  —  —  2j.     An  equation  which  is  incompatible  with  the 
first.     Hence  the  surface  has  no  centre. 

The  cubic  /  —  iS>r2  +  8ij=  O;  or  s*—i  8^  +  81  =  o, 


which   gives   two  roots  equal  to  9.     The  surface  is  therefore  a  para 
boloid  of  revolution 


—  2 


To  find  V,  we  first  determine  /,  m  and  n.     For  these  we  have  the 
equations 

4/—   ^  —  272  =  o 


—  o 


U-     U        •  7  2  2  ! 

which  Rive  /  =  —  ,  m  =  —    n  —  --  . 
3  3  3 

Therefore    (Eq.   M)      L'=  3  .-+  3  .-+-.•-  =  - 

and  2V  —  —  2L'  =—9. 

The  reduced  equation  of  the  surface  is  therefore 


Ex.  5.     2A70+2B'^v+2C'.r^'4-2Av.v  +  2B'jv4-2C"2;  —  D. 

The  cubic  is 

s«-(A'2  +  B'2  f-C'2)s-2A'B'C'  =  o. 

The  surface  is  a  hyperboloid  if  A',  B'  and  C'  are  all  different  from 
o.  If  A'B'C  is  of  the  same  sign  as  F  in  the  reduced  equation  the 
cubic  will  have  two  roots  of  the  same  sign  as  F  and  the  surface  will 
be  a  hyperboloid  of  one  sheet.  In  the  opposite  case  it  would  be  a  hyper 
boloid  of  two  sheets. 

If  A'  =  o  the  cubic  becomes 

j.2_(B''24-C'*)  =o,  whose  roots  are  of  different  signs.  Hence 
the  surface  2H'zx  +  2C'xy-\-  2A".r+  2B'V  +  2Cr'z  =  o  is  a  hyperbolic 
paraboloid. 

Ex.    6.  jc2+_y24-9s2  +  6r0  —  6xz  —  2~\y+  2.v—  42  —  o. 

The  equations  of  the  centre  are  incompatible  and  the  terms  of  the 


O.Y   SOLID    GEOMETRY. 


99 


second  degree  form  a  perfect  square,  hence  the  surface  is  a  parabolic 
cylinder. 

EXAMPLES. 

164.   i.  Find  the  nature  of  the  surfaces  represented  by  the  follow 
ing  equations. 

( i ).  i  i~v2  -f  5_y2  4-  2z2  —  2o\'z  +  4zx  4-  1 6xv  4-  2 2.v  +  1 6 v  4-  42  4- 1 1  =  o. 

(  2  ).     .X2  +/  4-  S*  4-  2F£  +  2X2  4-  2AJ  —  I  O,V—  I OV  —IOZ-\-2$  =  O. 

( 3 ).   3 jc2 — 372  —  1 2J2T  +i2zx  +  Sxv — 6x — 6>>  +  3^  =  0. 
(4).  4.i-2  +  QV2  +  97s2—  1 60jt-  +  54^  =36. 

(5).    3X*+2f—2XZ  \-  4VZ  —  4.V  —  80  —  8  =  O. 

2.  The  equation  7^-24-8/  +  4s2— 7^0—  nzx—  jxy  =  ^  represents 
a  hyperboloid  of  one  sheet. 

3.  The  equation  .r2 +>;2+32:2+  yz+zx+xy—'jx  —  1 4>'— 252  =  12—  d 
represents  an  ellipsoid,  a  point,  or  an  imaginary  surface  according 
as  d  is  <  =   >  67. 

4.  The  equation  _v24-.v84-28  -i-yz+zx  +  xy  =  a*  represents  an  oblate 
spheroid. 

5.  Find  the  nature  of  the  surface  (y— z)*+  (z— xY  +  (x— v)2  =  a~. 

6.  Find  the  nature  of  the  surface yz  \  zx  +  xy  —  #2. 

7.  ^a-2  +  4v'24-9s24-  I2yz  +  6zx  +  4xy+i4x+  16^+2404-47  =  o   re 
presents  an  elliptic,  a  parabolic  or  a  hyperbolic  cylinder  according  as 
a  >  =  <  i. 


CHAPTER   XII. 
PROBLEMS   OF    LOCI. 

165.  PROB.  I.  To  find  the  surface  of  revolution  generated  by  a  right 
line  turning  around  a  fixed  axis  which  it  does  not  intersect. 

Let  the  fixed  line  be  the  axis  of  z  and  let  the  shortest  distance  a 
from  the  revolving  line  to  the  axis  of  z  lie  along  the  axis  of  x  in  the 
original  position  of  this  line  so  that  its  equation  is  x  —  a,  y  =  mz. 

Then  the  equation  of  the  surface  is 


or 


The  hyperboloid  of  revolution  of  one  sheet. 

Prob.  2.  To  find  the  locus  of  a  point  whose  shortest  distances  from  two 
given  non-intersecting,  non-parallel  straight  lines  are  equal. 

Take  the  axis  of  z  along  the  shortest  distance  between  the  two 
lines,  the  plane  xy  perpendicular  to  z  at  the  middle  point  of  this 
distance  2c,  and  the  axes  of  x  and  y  bisecting  the  angles  between  the 
projections  of  the  line  on  their  plane.  Then  the  equation  of  the  lines 
will  be 

z  =  c      \       z  —  —c       ) 
y  —  mx  f  '  y  —  —  mx  } 

(  y  —  mxY  xo      (  V  +  mxV 

and  we  have  (z  —  cY  +  —  -  rL=  (*  +  0  +  ~  ~- 

I  +  m~  i  +  ///8 

or 

cz(i  +  ?n*)  -f  mxy  =  o,  a  hyperbolic  paraboloid  since  it  has  no 
centre  and  its  term  of  second  degree  breaks  up  into  two  real  factors. 

Prob.  3.    7  wo  planes   mutually  perpendicular,    contain   each  a  fixed 
straight  line.     To  find  the  surface  generated  by  their  line  of  intersection. 
Take  the  axes  as  in  Prob.  2.    Then  the  equations  of  the  planes  are 

100 


NOTES   ON  SOLID   GEOMETRY.  IOI 

K(z— c)  +y— mx  —  o  ;   (i)    K'(z  +  t)+y  +  Mx  =  G.   (2) 
The  condition  of  perpendicularity  of  these  planes  is 

KK'+i—  m*  —  o,  and  eliminating  K  +  K' between  this  equation 
and  equations  (i)  and  (2)  we  have 

y  _ /;/V  +  ( i  _  m*Y  —  ( i  -  m*y 
which  represents  a  hyperboloid  of  one  sheet. 

Prob.  4.  To  find  the  surface  generated  by  a  right*  line  which  always 
meets  three  fixed  right  lines  no  two  of  which  are  in  the^santy  fflantj  *++ 

For  greatest  simplicity  take  the  origin  at  the  Centre  of  &  paraljek)-- 
piped,  and  let  its  faces  be  at  the  distances  a,  &,  d  respectively  from- 
the  coordinate  planes yz,  xz,  and  xy.  Then  take  three  edges  of  this 
parallelopipedon  as  the  three  fixed  lines  fulfilling  the  conditions. 


Assume  for  the  equations  of  the  movable  line 

*^.m2Z* -•_=*.>      (4) 
cos   a       cos  ft       cos   y 

The  conditions  that  the  line  (4)  shall  meet  the  lines  (i)  (2)  and 
(3)  are  respectively 

y'  —  b  _   z  +c       z'  —  c  _  x  +  a    x  —  a  _  y'  -f  b 
cos  ft       cos  y '  cos  y  ~  cos   a  '  cos  a  ~~~  cos  ft  ' 

Eliminate  the  arbitrages  a,  ft,  y  by  multiplying  the  equations  to 
gether,  and  we  have  for  the  surface 


or  reducing 

ayz  +  bzx  +  cxy  +  abc  —  o, 

which   the   discriminating  cubic   shows  to   be  a  hyperboloid  of  one 
sheet.      The  same  surface  will  be  generated  by  a  straight  line  resting 

x  =•  a       }      y  •=  —  b  }      x  =  —  a 

on  the  other  three  edges  >  ,  -  ,  . 

z  =  —  C  ]      z  —  c       \      y  =  o 

Prob.    5.    To  find  the   surface  generated  by  a  right  line  which  alw  ns 
9* 


102  NOTES   ON  SOLID    GEOMETRY. 

meets  three  fixed  right  lines,  no  two  of  which  are  in  the  same  plane,  but  all 
of  which  are  parallel  to  the  same  plane. 

Take  one  of  the  fixed  lines  as  the  axis  of  x,  and  then  the  other  two 
parallel  to  the  plane  of  A:;'.     Then  their  equations  are 


Now,  the  equations  of  a  moving  line  meeting  lines  (i)  and  (2)  are 
-    .          ,-  (4)  (/  and  k  arbitrary),  and  the  condition  that  this 
jine  shall  also  meet  (.3)  is  Ic  —  mk  (c—b], 
arid  'eliminating  therV  and  k  by  means  of  equation  (4),  we  have 

cy        mx(c—b) 

2  z  —  b 

or  cyz  -\-m(b  —  c)xz  —  ( by  —  o, 

a  hyperbolic  paraboloid,  as  its  equation  shows  no   centre,  and  the 
terms  of  the  second  degree  break  up  into  two  real  factors. 

Prob.  6.   To  find  the  surface  generated  by  a  right  line  which  meets 
two  fixed  right  lines,  and  is  always  parallel  to  a  fixed  plane. 

Since  the  two  fixed  lines  must  meet  the  fixed  plane,  we  can  take 


-y      ——" 

Z 


1  1  (i),  y  [  (2),  as  in  2,  as  the  fixed  lines,  and  the 

=  c      \  z  =•  —  c      } 


plane  yz  as  the  fixed  plane. 

Then  the  equation  of  the  moving  line  parallel  toyz 


is  Z  (3),  /,  A  and  k  arbitrary. 

The  conditions  that  this  line  shall  meet  the  lines  (i)  and  (2) 

mk  =  k+/> 
-mk=-lc+t\ 

or  mk  =  Ic  and/>  =  o  ; 

or  eliminating  /,  k,  and  /, 

y 

mx  =  c  --  ; 
z 

or  mxz  =  cy,  a  hyperbolic  paraboloid. 


NOTES  ON  SOLID   GEOMETRY.  Ic>3 

Prob.  7.  Two  finite  non  -  intersecting  non-faralld  right  lines  are 
divided  each  into  the  same  number  of  equal  parts  ;  to  find  the  surface 
ivhich  is  the  locus  of  the  lines  joining  corresponding  points  of  divi 
sion. 

Let  the  line  which  joins  two  corresponding  extremities  of  the  given 
lines  be  the  axis  of  z  ;  let  the  axes  of  x  andjy  be  taken  parallel  to  the 
given  lines  and  the  plane  of  xy  be  halfway  between  them.  Let  the 
lengths  of  the  given  lines  be  a  and  b. 

Then  the  coordinates  of  two  corresponding  points  are 

z  =.  c,  x  —  ma,  y  =  o  ;  z  =  —  c,   x  =  o,  y  —  mb  ; 
and  the  equations  of  the  lines  joining  these  points  are 

—  +  ^-=1    ] 

ma       mb 

2X          Z 
---  =1 

?na        c 
whence  eliminating  m  the  equation  of  the  locus  is 


a  hyperbolic  paraboloid. 

Prob.  8.  To  find  the  locus  of  the  middle  points  of  chords  of  a  surface 
of  the  second  order  that  has  a  centre,  which  all  pass  through  a  given  fixed 
point. 

Take  the  given  point  for  the  origin  and  two  conjugate  diametral 
planes  which  pass  through  it  for  the  planes  of  zx  and  xy,  and  a  plane 
parallel  to  the  third  conjugate  plane  for  that  of>'2;  then  the  equation 
to  the  surface  will  be  of  the  form 

ax*  +  by1  +  cz1  +  2a"x  +/  —  o. 

Let  x  —  viz,  y  =  nz  be  the  equations  of  any  chord.  Combining 
these  with  the  equation  of  the  surface,  we  have 

(am*  +  bn^  +  c')zy  +  2ct'mz  +  d  —  o, 
in  which  the  values  of  z  belong  to  the  extremities  of  the  chord. 


104  VOTES  ON   SOLID   GEOMETRY. 

Therefore  the  z  of  its  middle  point  is 

a"m 


and  the  other  two  coordinates  of  the  middle  point  are 

x'=  mz't  (2)    y'  =  nz'.  (3) 
Hence  eliminating  m  and  n  the  required  equation  of  the  locus 


a  surface  of  the  second  order  similar  to  the  first,  and  passing  through 
its  centre  and  through  the  origin. 


CHAPTER  XIII. 
SOME    CURVES    OF    DOUBLE   CURVATURE. 

1 66.  To  find  the  equations  to  the  equable  spherical  spiral. 
Definition.     If  a  meridian  of  a  sphere  revolve  uniformly  about  its 

diameter  PP'  while  a  point  M  moves  uniformly  along  the  meridian 
from  P'  to  P",  so  as  to  describe  an  arc  equal  to  the  angle  through 
which  the  meridian  has  revolved,  the  locus  of  M  is  the  equable 
spherical  spiral. 

Taking  PP'  as  the  axis  of  z,  PAP'  the  initial  position  of  the  plane 
of  the  meridian  as  the  plane  of  xz,  the  equation  of  the  sphere  is 

ac*4y +*»  =  <!*. 

Let  MON  =  0,  AON  —  q),  then,  by  definition  6  =  cp, 
and  from  polar  coordinates 

x  =  a  cos  0  cos  cp,  r  —  a  cos  6  sin  cp ; 
/.  x  =  a  cos2  6,  y  =  a  cos  0  sin  0. 

Therefore  **+/  =  «»  cos2  0  (cos2  0  +  sin*  8)  =  ax. 

Hence  the  equations  of  the  spiral  are 

.r»+y  +  «»=  «2  (i)     #"  +  v*  =  ax;  (2) 

or  the  spiral   is  the   curve  of  intersection  of  the  sphere  and  a  right 
circular  cylinder  whose  diameter  is  the  radius  of  the  sphere. 
If  we  subtract  (2)  from  (i)  we  obtain 

z~  —  <?2  —ax  (3)  a  parabolic  cylinder. 

And  the  equations  (2)  and  (3)  also  represent  the  curve,  which  is 
therefore  also  the  intersection  of  a  right  circular  and  right  parabolic 
cylinder  at  right  angles  to  each  other. 

167.  To  find  the  equations  to  a  spherical  ellipse. 

Definition.     The  spherical  ellipse  is  a  curve  traced  on  the  surface 

105 


I06  A'OTES   ON   SOLID   GEOMETRY. 

of  a  sphere  such  that   the  sum   of  the  distances  of  any  point  on  it 
from  two  fixed  points  on  the  sphere  is  constant. 

Let  S,  H  be  the  t\vo  fixed  points  on  the  surface  of  the  sphere 
whose  radius  is  ;-,  C  the  middle  point  of  the  arc  of  the  great  circle 
which  joins  them.  If  P  be  any  point  of  the  spherical  ellipse,  SP  and 
HP  arcs  of  great  circles,  then 

SP-fHP  =  2a  =  a  constant. 

Through  P  draw  PM,  an  arc  of  a  great  circle  perpendicular  to 
SH,  and  let  SH  =  2;/,  CM  =  y,  PM  =  0. 

Then,  in  the  right-angled  spherical  triangle  SPM  we  have 

cos  SP  =  cos  (y  +  q>)  cos  0. 
And  in  the  triangle  HPM 

cos  PH  =  cos       —  f     cos  6. 


Now,  cos  SP  +  cos 

UTD                /SP+HP\         /SP-HPX 

HP  —  •?  PO«J    1                           I  rns   1 

V          2           /  °  f  \           2           ) 

/SP-HPX 
=  2  cos  a  cos  1  —           —  ) 

/SP  +  HPX          /SP-HPX 

And  cos  HP      cos 

Therefore, 
rns    f 

X           2           / 

=  2  sin 
'  SP-HPX        cosy  cos  cp 

\           2           J 

/SP-HPX 

*A     (  "-2         J 

COS  0 

2        J  cos  a 

/SP-HPN  sin  Y  sin  q>  cos  6 

sin                   -  V  ?B A    .  — . 

V        2        /  sin  «r 


Squaring  and  addini 


cos2  Y                    9  /)        s'°2  V    •   2  2/3 
—  cos   «  cos   c'  H — ; sin2  <z>  cos1  Cr-xsi; 

cos   r.f  sin-  (y 

or  if  we  transform  from  polar  to  rectangular  coordinates 

o  •       o 

cos* 


c 

+  -r-r^y  =  ^  (0  • 

' 


This  equation  and  the  equation  of  the  sphere 
A-9  +.!•*  +  **  nr  r2     (2) 

determine  the  spherical  ellipse,  as  the  intersection  of  a  right  elliptic 
cylinder  and  the  sphere. 


NOTES  ON   SOLID   GEOMETRY. 


107 


1 68.    To  find  the  equations  to  the  helix. 

Definition.  Whilst  the  rectangle  ABCM  revolves  uniformly  about 
its  side  AB,  so  that  the  parallel  side  CM  generates  the  surface  of  a 
right  circular  cylinder,  the  point  P  moves  uniformly  along  CM,  and 
generates  a  curve  called  a  helix. 

Let  AB  be  the  axis  of  z,  and  when  the  rectangle  is  in  the  plane 
xz  let  P  and  M  both  be  at  D  on  the  axis  of  x,  and  let  the  velocity  of 
P  =  n  times  the  velocity  of  M. 

.-.  PM  =  ;/.  arc. DM. 

Also  let  AN  =  jv,  NM  —  y,  PM  =  z  be  the  coordinates  of  P,  and 
AM  =  a  the  radius  of  the  circular  base  of  the  cylinder  in  the  plane  xy. 

.'.  z  =  na  cos"1    —  ,   and    j'2  +  x^  —  <r     (i) 

are  the  required  equations  of  the  helix. 

Or  we  may  represent  the  curve  by  the  two  equations 

z  —  na  cos"1  — .  z  —  na  sin"1  —     (2): 
a  a 

or  the  same  in  the  forms 

z  z 

x  =  a  cos  —  ,    y  =  a  sin  — ,     n)     and 
na  na 

z  /  z  \  z  /  z 

since  cos  —  —  cos    2m  7t  -\ and  sin  - —  =  sin      2m  n-\ 

na  \  na  J  na  \  na 

the  same  values  of  x  and  y  correspond  to  an  infinite  number  of 
values  of  z.  The  equations  (i)  (2)  and  (3)  show  that  the  projec 
tions  of  the  helix  on  the  planes  xz,  and  yz  give  the  curve  of  sines, 
and  the  projection  on  xy  is  the  circle. 


PLATE  T 


PLATE  II 


7  DAY  USE 

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